MATH/STAT. 


T.  SUNDARA  ROW'S 

Geometric  Exercises  in 
Paper  Folding 


Edited  and  Revised  by 
WOOSTER  WOODRUFF  BEMAN 

PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY  OF  MICHIGAK 

and 
DAVID  EUGENE  SMITH 

PROFESSOR  OF  MATHEMATICS  IN  TEACHERS1  COLLEGE 
OF  COLUMBIA  UNIVERSITY 


WITH  87  ILLUSTRATIONS 


THIRD  EDITION 


CHICAGO     :::     LONDON 
THE     OPEN     COURT    PUBLISHING    COMPANY 

1917 


& 


» 


XC? 

4255 


COPYRIGHT  BY 

THE  OPEN  COURT  PUBLISHING  Co, 
1901 


PRINTED  IN  THE   UNITED    STATES  OF    AMERICA 


hn 

HATH 


EDITORS'  PREFACE. 

OUR  attention  was  first  attracted  to  Sundara  Row's  Geomet 
rical  Exercises  in  Paper  Folding  by  a  reference  in  Klein's  Vor- 
lesungen  iiber  ausgezucihlte  Fragen  der  Elementargeometrie. 
An  examination  of  the  book,  obtained  after  many  vexatious  delays, 
convinced  us  of  its  undoubted  merits  and  of  its  probable  value  to 
American  teachers  and  students  of  geometry.  Accordingly  we 
sought  permission  of  the  author  to  bring  out  an  edition  in  this 
country,  wnich  permission  was  most  generously  granted. 

The  purpose  of  the  book  is  so  fully  set  forth  in  the  author's 
introduction  that  we  need  only  to  say  that  it  is  sure  to  prove  of 
interest  to  every  wide-awake  teacher  of  geometry  from  the  graded 
school  to  the  college.  The  methods  are  so  novel  and  the  results 
so  easily  reached  that  they  cannot  fail  to  awaken  enthusiasm. 

Our  work  as  editors  in  this  revision  has  been  confined  to  some 
slight  modifications  of  the  proofs,  some  additions  in  the  way  of 
references,  and  the  insertion  of  a  considerable  number  of  half-tone 
reproductions  of  actual  photographs  instead  of  the  line-drawings 
of  the  original. 

W.  W.  BEMAN. 
D.  E.  SMITH. 


OK02 


CONTENTS. 

PAGE 

Introduction m        •     •  vn 

I.   The  Square ' i 

II.  The  Equilateral  Triangle 9 

III.  Squares  and  Rectangles 14 

IV.  The  Pentagon 3° 

V.  The  Hexagon , 35 

VI.  The  Octagon 39 

VII.  The  Nonagon 45 

VIII.  The  Decagon  and  the  Dodecagon 47 

IX.  The  Pentedecagon 5° 

X.   Series 52 

XI.  Polygons 67 

XII.   General  Principles 82 

XIII.  The  Conic  Sections. 

Section     i.  The  Circle 102 

Section    n.   The  Parabola 115 

Section  in.   The  Ellipse 121 

Section  iv.   The  Hyperbola 126 

XIV.  Miscellaneous  Curves 131 


INTRODUCTION. 

THE  idea  of  this  book  was  suggested  to  me  by 
Kindergarten  Gift  No.  VIII. — Paper-folding.  The 
gift  consists  of  two  hundred  variously  colored  squares 
of  paper,  a  folder,  and  diagrams  and  instructions  for 
folding.  The  paper  is  colored  and  glazed  on  one  side. 
The  paper  may,  however,  be  of  self-color,  alike  on 
both  sides.  In  fact,  any  paper  of  moderate  thickness 
will  answer  the  purpose,  but  colored  paper  shows  the 
creases  better,  and  is  more  attractive.  The  kinder 
garten  gift  is  sold  by  any  dealers  in  school  supplies  ; 
but  colored  paper  of  both  sorts  can  be  had  from  sta 
tionery  dealers.  Any  sheet  of  paper  can  be  cut  into 
a  square  as  explained  in  the  opening  articles  of  this 
book,  but  it  is  neat  and  convenient  to  have  the  squares 
ready  cut. 

2.  These  txercises   do   not  require   mathematical 
instruments,  the  only  things  necessary  being  a  pen 
knife   and  scraps  of  paper,  the  latter  being  used  for 
setting  off  equal  lengths.    The  squares  are  themselves 
simple  substitutes  for  a  straight  edge  and  a  T  square. 

3.  In  paper-folding   several   important   geometric 
processes  can  be  effected  much  more  easily  than  with 


viii  INTR  OD  UC  TION. 

a  pair  of  compasses  and  ruler,  the  only  instruments 
the  use  of  which  is  sanctioned  in  Euclidean  geom 
etry  ;  for  example,  to  divide  straight  lines  and  angles 
into  two  or  more  equal  parts,  to  draw  perpendiculars 
and  parallels  to  straight  lines.  It  is,  however,  not 
possible  in  paper-folding  to  describe  a  circle,  but  a 
number  of  points  on  a  circle,  as  well  as  other  curves, 
may  be  obtained  by  other  methods.  These  exercises 
do  not  consist  merely  of  drawing  geometric  figures 
involving  straight  lines  in  the  ordinary  way,  and  fold 
ing  upon  them,  but  they  require  an  intelligent  appli 
cation  of  the  simple  processes  peculiarly  adapted  to 
paper-folding.  This  will  be  apparent  at  the  very  com 
mencement  of  this  book. 

4.  The  use  of  the  kindergarten  gifts  not  only  affords 
interesting  occupations  to  boys  and  girls,  but  also 
prepares  their  minds  for  the  appreciation  of  science 
and  art.  Conversely  the  teaching  of  science  and  art 
later  on  can  be  made  interesting  and  based  upon 
proper  foundations  by  reference  to  kindergarten  occu 
pations.  This  is  particularly  the  case  with  geometry, 
which  forms  the  basis  of  every  science  and  art.  The 
teaching  of  plane  geometry  in  schools  can  be  made 
very  interesting  by  the  free  use  of  the  kindergarten 
gifts.  It  would  be  perfectly  legitimate  to  require  pu 
pils  to  fold  the  diagrams  with  paper.  This  would 
give  them  neat  and  accurate  figures,  and  impress  the 
truth  of  the  propositions  forcibly  on  their  minds.  It 
would  not  be  necessary  to  take  any  statement  on  trust. 


INTR  OD  UC  TION.  ix 

But  what  is  now  realised  by  the  imagination  and  ideal 
isation  of  clumsy  figures  can  be  seen  in  the  concrete. 
A  fallacy  like  the  following  would  be  impossible. 

5.  To  prove  that  every  triangle  is  isosceles.  Let 
ABC,  Fig.  1,  be  any  triangle.  Bisect  AB  in  Z,  and 
through  Z  draw  ZO  perpendicular  to  AB.  Bisect  the 
angle  ACB  by  CO.  * 


A  2  B 

Fig.  i. 

(1)  If  CO  and  ZO  do  not  meet,  they  are  parallel. 
Therefore  CO  is  at  right  angles  to  AB.     Therefore 
AC^BC. 

(2)  If  CO  and  ZO  do   meet,  let   them   meet  in  O. 
Draw  OX  perpendicular  to  BC  and  OY perpendicular 
to  AC.     Join   OAy  OB.     By  Euclid  I,  26  (B.  and  S., 
§  88,  cor.  7)*  the  triangles    YOC  and  XOC  are   con- 

*  These  references  are  to  Beman  and  Smith's  New  Plane  and  Solid  Geom 
etry,  Boston,  Ginn  &  Co.,  1899. 


x  INTRODUCTION. 

gruent;  also  by  Euclid  I,  47  and  I,  8  (B.  and  S.,  § 
156  and  §  79)  the  triangles  AOY  and  BOX  are  con 
gruent.  Therefore 

AY+  YC=BX+XC, 

i.e.,  AC^BC. 

Fig.  2  shows  by  paper- folding  that,  whatever  tri 
angle  be  taken,  CO  and  ZO  cannot  meet  within  the 
triangle. 


Fig.  2. 

O  is  the  mid-point  of  the  arc  A  OB  of  the  circle 
which  circumscribes  the  triangle  ABC. 

6.  Paper-folding  is  not  quite  foreign  to  us.  Fold 
ing  paper  squares  into  natural  objects — a  boat,  double 


IN  TR  OD  UC  T1ON.  xi 

boat,  ink  bottle,  cup-plate,  etc.,  is  well  known,  as 
also  the  cutting  of  paper  in  symmetric  forms  for  pur 
poses  of  decoration.  In  writing  Sanskrit  and  Mah- 
rati,  the  paper  is  folded  vertically  or  horizontally  to 
keep  the  lines  and  columns  straight.  In  copying  let 
ters  in  public  offices  an  even  margin  is  secured  by  fold 
ing  the  paper  vertically.  Rectangular  pieces  of  paper 
folded  double  have  generally  been  used  for  writing, 
and  before  the  introduction  of  machine-cut  letter  pa 
per  and  envelopes  of  various  sizes,  sheets  of  convenient 
size  were  cut  by  folding  and  tearing  larger  sheets,  and 
the  second  half  of  the  paper  was  folded  into  an  envel 
ope  inclosing  the  first  half.  This  latter  process  saved 
paper  and  had  the  obvious  advantage  of  securing  the 
post  marks  on  the  paper  written  upon.  Paper-folding 
has  been  resorted  to  in  teaching  the  Xlth  Book  of 
Euclid,  which  deals  with  figures  of  three  dimensions.* 
But  it  has  seldom  been  used  in  respect  of  plane  fig 
ures. 

7.  I  have  attempted  not  to  write  a  complete  trea 
tise  or  text-book  on  geometry,  but  to  show  how  reg 
ular  polygons,  circles  and  other  curves  can  be  folded 
or  pricked  on  paper.  I  have  taken  the  opportunity  to 
introduce  to  the  reader  some  well  known  problems  of 
ancient  and  modern  geometry,  and  to  show  how  alge 
bra  and  trigonometry  may  be  advantageously  applied 
to  geometry,  so  as  to  elucidate  each  of  the  subjects 
which  are  usually  kept  in  separate  pigeon-holes. 

*  See  especially  Beman  and  Smith's  New  Plane  and  Solid  Geometry,  p.  287. 


xii  INTR  OD  UC  TION. 

8.  The  first  nine  chapters  deal  with  the  folding  of 
the  regular  polygons   treated  in  the  first  four  books  of 
Euclid,  and  of  the  nonagon.    The  paper  square  of  the 
kindergarten   has   been  taken  as  the  foundation,  and 
the   other   regular   polygons    have    been   worked    out 
thereon.   Chapter  I  shows  how  the  fundamental  square 
is  to  be  cut  and  how  it  can  be  folded  into  equal  right- 
angled   isosceles   triangles   and   squares.      Chapter  II 
deals  with  the  equilateral  triangle  described  on  one  of 
the  sides  of  the  square.      Chapter  III  is  devoted  to 
the  Pythagorean  theorem  (B.  and  S.,  §  156)  and  the 
propositions  of  the  second  book  of  Euclid  and  certain 
puzzles  connected  therewith.      It  is  also  shown  how  a 
right-angled  triangle  with  a  given  altitude  can  be  de 
scribed  on  a  given  base.     This  is  tantamount  to  find 
ing  points  on  a  circle  with  a  given  diameter. 

9.  Chapter  X  deals  with  the  arithmetic,  geometric, 
and  harmonic  progressions  and  the  summation  of  cer 
tain  arithmetic  series.    In  treating  of  the  progressions, 
lines  whose  lengths  form  a  progressive  series   are  ob 
tained.      A  rectangular  piece  of  paper  chequered  into 
squares  exemplifies  an  arithmetic  series.    For  the  geo 
metric  the  properties  of  the  right-angled  triangle,  that 
the  altitude  from  the  right  angle  is  a  mean  propor 
tional  between   the  segments  of  the  hypotenuse  (B. 
and  S.,  §  270),  and  that  either  side  is  a  mean  propor 
tional  between  its  projection  on  the  hypotenuse  and 
the  hypotenuse,  are  made  use  of.      In  this  connexion 
the  Delian  problem  of  duplicating  a  cube  has  been 


INTRODUCTION.  xiii 

explained.*  In  treating  of  harmonic  progression,  the 
fact  that  the  bisectors  of  an  interior  and  correspond 
ing  exterior  angle  of  a  triangle  divide  the  opposite 
side  in  the  ratio  of  the  other  sides  of  the  triangle  (B. 
and  S.,  §  249)  has  been  used.  This  affords  an  inter 
esting  method  of  graphically  explaining  systems  in 
involution.  The  sums  of  the  natural  numbers  and  of 
their  cubes  have  been  obtained  graphically,  and  the 
sums  of  certain  other  series  have  been  deduced  there 
from. 

10.  Chapter  XI  deals  with  the  general  theory  of 
regular  polygons,  and  the  calculation  of  the  numerical 
value  of  7t.     The  propositions  in  this  chapter  are  very 
interesting. 

11.  Chapter  XII   explains  certain  general  princi 
ples,  which  have  been  made  use  of  in  the  preceding 
chapters, — congruence,   symmetry,   and   similarity  of 
figures,  concurrence  of  straight  lines,  and  collinearity 
of  points  are  touched  upon. 

12.  Chapters  XIII  and  XIV  deal  with  the  conic 
sections    and    other    interesting    curves.      As   regards 
the  circle,  its  harmonic  properties  among  others  are 
treated.   The  theories  of  inversion  and  co-axial  circles 
are   also  explained.     As    regards    other   curves  it  is 
shown  how  they  can  be  marked  on   paper  by  paper- 
folding.     The  history  of  some  of  the  curves  is  given, 
and  it  is  shown  how  they  were  utilised  in  the  solution 

*See  Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of  Ele 
mentary  Geometry,  Boston,  1897;  also  their  translation  of  Fink's  History  of 
Mathematics,  Chicago,  The  Open  Court  Pub.  Co.,  1900. 


xi  v  INTR  OD  UC  TION. 

of  the  classical  problems,  to  find  two  geometric  means 
between  two  given  lines,  and  to  trisect  a  given  recti 
lineal  angle.  Although  the  investigation  of  the  prop 
erties  of  the  curves  involves  a  knowledge  of  advanced 
mathematics,  their  genesis  is  easily  understood  and  is 
interesting. 

13.  I  have  sought  not  only  to  aid  the  teaching  of 
geometry  in  schools  and  colleges,  but  also  to  afford 
mathematical  recreation  to  young  and  old,  in  an  at 
tractive  and  cheap  form.  "Old  boys"  like  mysell 
may  find  the  book  useful  to  revive  their  old  lessons, 
and  to  have  a  peep  into  modern  developments  which, 
although  very  interesting  and  instructive,  have  been 
ignored  by  university  teachers. 

T.  SUNDARA  Row. 
MADRAS,  INDIA,  1893. 


I.   THE  SQUARE. 

1.  The  upper  side  of  a  piece  of  paper  lying  flat 
upon  a  table  is  a  plane  surface,  and  so  is  the  lower 
side  which  is  in  contact  with  the  table. 

2.  The  two  surfaces  are  separated  by  the  material 
of  the  paper.   The  material  being  very  thin,  the  other 
sides  of  the  paper  do  not  present  appreciably  broad 
surfaces,  and  the  edges  of  the  paper  are  practically 
lines.     The  two  surfaces  though  distinct  are  insepa 
rable  from  each  other. 

3.  Look  at  the  irregularly  shaped  piece  of  paper 
shown  in  Fig.  3,  and  at  this  page  which  is  rectangu 
lar.      Let  us  try  and  shape  the  former  paper  like  the 
latter. 

4  Place  the  irregularly  shaped  piece  of  paper 
upon  the  table,  and  fold  it  flat  upon  itself.  Let  X'  X 
be  the  crease  thus  formed.  It  is  straight.  Now  pass 
a  knife  along  the  fold  und  separate  the  smaller  piece. 
We  thus  obtain  one  straight  edge. 

5.  Fold  the  paper  again  as  before  along  BY,  so 
that  the  edge  X' X  is  doubled  upon  itself.  Unfolding 
the  paper,  we  see  that  the  crease  ^Kis  at  right  angles 
to  the  edge  X'X.  It  is  evident  by  superposition  that 


GEOMETRIC  EXERCISES 


the  angle  YBX'  equals  the  angle  XBY,  and  that  each 
of  these  angles  equals  an  angle  of  the  page.    Now  pass 


Fig.  3- 


a  knife  as  before  along  the  second  fold  and  remove 
the  smaller  piece. 


IN  PAPER  FOLDING 


6.  Repeat  the  above  process  and  obtain  the  edges 
CD  and  DA.  It  is  evident  by  superposition  that  the 
angles  at  A,  B,  C,  £>,  are  right  angles,  equal  to  one 
another,  and  that  the  sides  BC,  CD  are  respectively 


Fig.  4. 

equal  to  DA,  AB.     This  piece  of  paper   (Fig.  3)  is 
similar  in  shape  to  the  page. 

7.  It  can  be  made  equal  in  size  to  the  page  by 
taking  a  larger  piece  of  paper  and  measuring  off  AB 
and  BC  equal  to  the  sides  of  the  latter. 


4  GEOMETRIC  EXERCISES 

8.  A   figure   like   this   is   called   a   rectangle.      By 
superposition  it  is  proved  that  (1)  the  four  angles  are 
right  angles  and  all  equal,  (2)  the  four  sides  are  not 
all  equal.  (3)  but  the  two  long  sides  are  equal,  and  so 
also  are  the  two  short  sides. 

9.  Now  take  a  rectangular  piece  of  paper,  A'B'  CD, 
and  fold  it  obliquely  so  that  one  of  the  short  sides,  CD, 


Fig.  5- 

falls  upon  one  of  the  longer  sides,  DA',  as  in  Fig.  4. 
Then  fold  and  remove  the  portion  A' B'BA  which 
overlaps.  Unfolding  the  sheet,  we  find  that  ABCD 
is  now  square,  i.  e. ,  its  four  angles  are  right  angles, 
and  all  its  sides  are  equal. 

10.  The  crease  which  passes  through  a  pair  of  th& 


IN  PAPER  FOLDING  5 

opposite  corners  B,  D,  is  a  diagonal  of  the  square. 
One  other  diagonal  is  obtained  by  folding  the  square 
through  the  other  pair  of  corners  as  in  Fig.  5. 

n.  We  see  that  the  diagonals  are  at  right  angles 
to  each  other,  and  that  they  bisect  each  other. 

12.  The  point  of  intersection  of  the  diagonals  is 
called  the  center  of  the  square. 


Fig.  6. 

• 

13.  Each  diagonal  divides  the  square  into  two  con 
gruent  right-angled  isosceles  triangles,  whose  vertices 
are  at  opposite  corners. 

14.  The  two  diagonals  together  divide  the  square 
into  four  congruent  right-angled    isosceles  triangles, 
whose  vertices  are  at  the  center  of  the  square. 


6  GEOMETRIC  EXERCISES 

15.  Now  fold  again,  as  in  Fig.  6,  laying  one  side 
of  the  square  upon  its  opposite  side.  We  get  a  crease 
which  passes  through  the  center  of  the  square.  It  is 
at  right  angles  to  the  other  sides  and  (1)  bisects  them; 
(2)  it  is  also  parallel  to  the  first  two  sides ;  (3)  it  is 
itself  bisected  at  the  center ;  (4)  it  divides  the  square 


Fig.  7. 


into  two  congruent  rectangles,  which  are,  therefore, 
each  half  of  it;  (5)  each  of  these  rectangles  is  equal 
to  one  of  the  triangles  into  which  either  diagonal 
divides  the  square. 

16.  Let  us  fold  the  square  again,   laying  the  re 
maining  two  sides  one  upon  the  other.     The  crease 


IN  PAPER  FOLDING  7 

now  obtained  and  the  one  referred  to  in   §  15  divide 
the  square  into  four  congruent  squares. 

17.  Folding  again  through  the  corners  of  the 
smaller  squares  which  are  at  the  centers  of  the  sides 
of  the  larger  square,  we  obtain  a  square  which  is  in 
scribed  in  the  latter.  (Fig.  7.) 


Fig.  8. 

18.  This  square  is  half  the  larger  square,  and  has 
the  same  center. 

19.  By  joining  the  mid-points  of  the  sides  of  the 
inner  square,  we  obtain  a  square  which  is  one-fourth 
of  the  original  square  (Fig.  8).    By  repeating  the  pro 
cess,  we  can  obtain  any  number  of  squares  which  are 
to  one  another  as 


8  GEOMETRIC  EXERCISES 

1  l  1  JL  i  L  :  L 

"2*  "4"'  ~8~'  16'  etc''  or  "2"'  2^'  jp'  24'  '  '  '  * 

Each  square  is  half  of  the  next  larger  square, 
i.  e.,  the  four  triangles  cut  from  each  square  are  to 
gether  equal  to  half  of  it.  The  sums  of  all  these  tri 
angles  increased  to  any  number  cannot  exceed  the 
original  square,  and  they  must  eventually  absorb  the 
whole  of  it. 

Therefore  —  -f  ^  +  ^  +  etc-  to  infinity  =  1. 

20.  The  center  of  the  square  is  the  center  of  its 
circumscribed  and  inscribed  circles.    The  latter  circle 
touches  the  sides  at   their   mid-points,   as  these  are 
nearer  to   the  center  than   any  other  points  on  the 
sides. 

21.  Any  crease  through  the  center  of  the  square 
divides  it  into  two  trapezoids  which  are  congruent.    A 
second  crease  through  the  center  at  right  angles  to 
the  first  divides  the  square  into  four  congruent  quadri 
laterals,  of  which  two  opposite  angles  are  right  angles. 
The  quadrilaterals  are  concyclic,  i.  e.,  the  vertices  of 
each  lie  in  a  circumference. 


II.   THE  EQUILATERAL  TRIANGLE. 

22.  Now  take  this  square  piece  of  paper  (Fig.  9), 
and  fold  it  double,  laying  two  opposite  edges  one  upon 
the  other.  We  obtain  a  crease  which  passes  through 


Fig.  9. 


the  mid-points  of  the  remaining  sides  and  is  at  right 
angles  to  those  sides.  Take  any  point  on  this  line, 
fold  through  it  and  the  two  corners  of  the  square  which 


io  GEOMETRIC  EXERCISES 

are  on  each  side  of  it.   We  thus  get  isosceles  triangles 
standing  on  a  side  of  the  square. 

23.  The  middle  line  divides  the  isosceles  triangle 
into  two  congruent  right-angled  triangles. 

24.  The  vertical  angle  is  bisected. 

25.  If  we  so  take  the  point  on  the  middle  line,  that 


Fig.  io. 

its  distances  from  two  corners  of  the  square  are  equal 
to  a  side  of  it,  we  shall  obtain  an  equilateral  triangle 
(Fig.  10).  This  point  is  easily  determined  by  turning 
the  base  AB  through  one  end  of  it,  over  A  A',  until  the 
other  end,  B,  rests  upon  the  middle  line,  as  at  C. 
26.  Fold  the  equilateral  triangle  by  laying  each 


IN  PAPER  FOLDING  n 

of  the  sides  upon  the  base.  We  thus  obtain  the 
three  altitudes  of  the  triangle,  viz.:  A  A',  BB',  CC, 
(Fig.  11). 

27.  Each  of  the  altitudes  divides  the  triangle  into 
two  congruent  right-angled  triangles. 

28.  They  bisect  the  sides  at  right  angles. 


Fig.  ix. 

29.  They  pass  through  a  common  point. 

30.  Let  the  altitudes  AA'  and    CC'   meet  in   O. 
Draw  BO  and  produce  it  to  meet  AC  in  B '.     BB' 
will  now  be  proved   to  be  the  third   altitude.      From 
the   triangles   C'OA   and    CO  A',    OC'=OA'.     From 
triangles  OC'B  and  A'OB,  ^OBC'^^A'BO.    Again 
from  triangles  ABB'  and  CB'B,    /_AB'B  =  /_BB C, 


12  GEOMETRIC  EXERCISES 

i.  e.,  each  of  them  is  a  right  angle.  That  is,  BOB'  is 
an  altitude  of  the  equilateral  triangle  ABC.  It  also 
bisects  AC  in  B' . 

31.  It  can  be  proved   as  above  that  OA,  OB,  and 
OC  are  equal,  and  that  OA' ,  OB',  and  OC'  are  also 
equal. 

32.  Circles  can  therefore  be  described  with  O  as  a 
center  and  passing   respectively  through  A,  B,  and  C 
and  through  A',  B' ',  and  C .    The  latter  circle  touches 
the  sides  of  the  triangle. 

33.  The  equilateral  triangle  ABC  is  divided  into 
six  congruent  right-angled  triangles  which  have  one 
set  of  their  equal  angles  at  O,  and  into  three  congru 
ent,  symmetric,  concyclic  quadrilaterals. 

34.  The  triangle  A  OC  is  double  the  triangle  A' OC; 
therefore,    AO  =  2OA'.      Similarly,    BO=r-2OL'   and 
CO  =  2OC'.      Hence  the  radius  of  the  circumscribed 
circle  of  triangle  ABC  is  twice  the  radius  of  the  in 
scribed  circle. 

35.  The  right  angle  A,  of  the  square,  is  trisected 
by  the  straight  lines  AO,  AC.      Angle  BAC=^\  of  a 
right  angle.      The  angles  C' A  O  and  OAB'  are  each  i 
of  a  right  angle.   Similarly  with  the  angles  at  B  and  C. 

36.  The  six  angles  at  O  are  each  £  of  a  right  angle. 

37.  Fold  through  A'B',  B' C' ,  and  C  A'  (Fig.  12). 
Then  A' B' C'  is  an  equilateral  triangle.     It  is  a  fourth 
of  the  triangle  ABC. 

38.  A'B',  B'C',  C'A'  are  each  parallel  to  AB,  BC, 
CA,  and  halves  of  them. 


JN  PAPER  FOLDING 


13 


39.  ACA'B'  is  a  rhombus.      So  are  C'BA'B'  and 
CB'C'A'. 

40.  A'B',  B'C',  C'A'  bisect  the  corresponding  alti 
tudes. 

41.  CC 


=  0.866.... 


Fig.  12. 

42.  The  A  ABC=  rectangle  of  AC  and  CC',  i.  e. 
X  \V*'AB  =  IVZ-AB?  =  0.433 ....  X^2- 

43.  The  angles  of   the  triangle  AC'C  are  in  the 
ratio  of  1:2:3,  and  its  sides  are  in  the  ratio  of  1/1 
:  1/3  :  1/4 . 


III.   SQUARES  AND  RECTANGLES. 

44.  Fold   the   given  square   as   in   Fig.    13.     This 
affords  the  well-known  proof  of  the  Pythagorean  the- 


Fig.  13. 

orem.    FGH  being  a  right-angled  triangle,  the  square 
on  FH  equals  the  sum  of  the  squares  on  FG  and  GH. 

uFA  +  u  DB  =  n  FC. 
It  is  easily  proved  that  FC  is  a  square,  and  that 


PAPER  FOLDING  15 

the  triangles  FGH,  HBC,  KDC,  and  FEK  are  con 
gruent. 

If  the  triangles  FGH  and  HBC  are  cut  off  from 
the  squares  FA   and  DB,  and   placed  upon  the  other 
two  triangles,  the  square  FHCK  is  made  up. 
IiAB  =  a,  GA=b,  and  FH=c,  then  a2  -f  6*  =  c*. 


Fig.  14. 

45.  Fold  the  given  square  as  in  Fig.  14.  Here  the 
rectangles  AF,  BG,  Cff,  and  DE  are  congruent,  as 
also  the  triangles  of  which  they  are  composed.  EFGH 
is  a  square  as  also  KLMN. 

Let  AK=a,  KB  =  b,  and  NK  =  c, 

then  a* +  #«  =^2,  j.  e.  uKLMN. 


i6 


GEOMETRIC  EXERCISES 


Now  square  ABCD  overlaps  the  square  KLMN 
by  the  four  triangles  AKN,  BLK,  CML,  and  DNM. 

But  these  four  triangles  are  together  equal  to  two 
of  the  rectangles,  i.  e.,  to  lab. 

Therefore  (a  -f  £)2  =  a2  +  P  -f  2ab. 

46.  EF=a  —  b,  auidn£FGJ?=(a  —  fi)*. 

The  square  EFGHis  less  than  the  square  KLMN 
by  the  four  triangles  FNKt  GKL,  HLM,  and  EMN. 

But  these  four  triangles  make  up  two  of  the  rect 
angles,  i.  e.,  Zab. 


Fig.  15. 

47.  The  square  ABCD  overlaps  the  square  EFGH 
by  the  four  rectangles  AF,  BG,  Clf,  and  DE. 

48.  In  Fig.  15,  the  square  ABCD  =  (a-\-  £)2,  and 


IN  PAPER  FOLDING  17 

the  square   EFGH—(a  —  <£)2.      Also   square   AKGN 
=  square    ELCM  =  a2.       Square    KBLF  =  square 

T   ft 


Squares  ABCD  and  EFGH  are  together  equal  to 
the  latter  four  squares  put  together,  or  to  twice  the 
square  AKGN  and  twice  the  square  KBLF,  that  is, 

(a  +  £)2  -f  (a  —  £)2  =  2a2  4-  2^2. 

D O N C 


Fig.  16. 


49.  In  Fig.  16  the  rectangle  PL  is  equal  to  (a-\-ft 
(«-*). 

Because  the  rectangle  EK  =  FM,  therefore  rect 
angle  PL  =  square  PK  —  square  AE,  i.  e.,  (a-\- 


50.  If  squares  be  described  about  the  diagonal  of 
a  given  square,  the  right  angle  at  one  corner  being 
common  to  them,  the  lines  which  join  this  corner  with 
the  mid-points  of  the  opposite  sides  of  the  given 


i8  GEOMETRIC  EXERCISES 

square  bisect  the  corresponding  sides  of  all  the  inner 
squares.  (Fig.  17.)  For  the  angles  which  these  lines 
make  with  the  diagonal  are  equal,  and  their  magni 
tude  is  constant  for  all  squares,  as  may  be  seen  by 


Fig.  17. 

superposition.     Therefore  the  mid-points  of  the  sides 
of  the  inner  squares  must  lie  on  these  lines. 

51.  ABCD  being  the  given  square  piece  of  paper 
(Fig.  18),  it  is  required  to  obtain  by  folding,  the  point 
^Tin  AB,  such  that  the  rectangle  AB-XB  is  equal  to 
the  square  on  AX, 

Double  BC  upon  itself  and  take  its  mid-point  E. 

Fold  through  E  and  A. 

Lay  EB  upon  EA  and  fold  so  as  to  get  EF,  and 
G  such  that  EG  =  EB. 

Take  AX=AG. 


IN  PAPER  FOLDING 


19 


Then  rectangle  AB-XB^AX*. 
Complete  the  rectangle  BCHX   and    the  square 
AXKL. 

Let  Xffcut  EA  in  M.     Take  FY=FB. 
Then  F£  =  FG  =  FY=XM  m&  XM=\AX. 


Fig.  18. 

Now,  because  BY  is  bisected  in  F  and  produced 
to  A, 

Y+FY'*  =  AF'i,  by  §  49> 

—  ^(92+^C2,  by  §  44. 
.-.  AB-AY=AG1J 


But  ^^2= 


20  GEOMETRIC  EXERCISES 

AY=XB. 


AB  is  said  to  be  divided  in  X  in  median  section.* 
Also 


i.  e.,  AB  is  also  divided  in  Y  in  median  section. 

52.  A  circle  can  be  described  with  F  as  a  center, 
its  circumference  passing  through  B,  G,  and  K      It 
will  touch  .ZL4   at  G,  because  FG  is  the  shortest  dis 
tance  from  F  to  the  line  EGA. 

53.  Since 


subtracting  BK  we  have 

rectangle  XKNY=  square  CHKP, 

i.  e.,  ^.AT-KA^^F2, 
i.  e.,  AX  is  divided  in  Kin  median  section. 

Similarly  j#Fis  divided  in  X  in  median  section. 


54.   ' 

CD-CP 


55.  Rectangles  Bff  and  YD  being  each  =  AB-XBy 
rectangle  ^F+  square  CK=AX^  =  AB-  XB. 


56.   Hence    rectangle    JfY=  rectangle  j^A",    i.  e., 


57.   Hence  rectangle  ^AT=  AX-XB—BX*. 


*  The  term  "  golden  section"  is  also  used.    See  Beman  and  Smith's  New 
Plane  and  Solid  Geometry,  p.  196. 


IN  PAPER  FOLDING  21 

58.  Let  AB  =  a,  XB  =  x. 

Then  (a  —  x^  =  ax,  by  §  51. 
0a  +  **=:3<i#,  by  §  54; 


and  #  =  .£(3  —  1/5). 


.-.  a  —  *  =  4-  (1/5  —  1)  =  *  X  0.6180.  . 

a 

...  (^_jc)2  =  ^(3  —  1/5)  =  ^2  X  0.3819. 

z 

The  rect.  BPKX 


=  a2  (1/5  —  2)  =  a2  X  0.2360 

~ 

4 


59.  In  the  language  of  proportion 

AB  :  ^^T=^^:  Xff.. 

The  straight  line  AB  is  said  to  be  divided  "in  ex 
treme  and  mean  ratio." 

60.  Let  AB  be  divided   in  X  in  median  section. 
Complete  the  rectangle  CBXH  (Fig.  19).      Bisect  the 
rectangle  by  the  line  MNO.      Find  the  point  N  by 
laying  XA   over  X  so  that  A   falls  on  MO,  and  fold 
through  XN,  NB,  and  7V^.      Then  BAN  is  an  isos- 


22 


GEOMETRIC  EXERCISES 


celes  triangle  having  its  angles  ABN  and  BNA  double 
the  angle  NAB. 

AX=XN=NB 


Fig.  19. 


and 


IN  PAPER  FOLD  IN  C 

.-.  AN=AB 

=  \  of  a  right  angle. 


61.  The  right  angle  at  A  can  be  divided  into  five 
equal  parts  as  in  Fig.  20.  Here  N'  is  found  as  in 
§60.  Then  fold^^'<2;  bisect  £QAB  by  folding, 


p'  Q  <~ 

Fig.  20. 

fold  over  the  diagonal  AC  and  thus  get  the  point 
Q',  /"• 

62.  To  describe  a  right-angled  triangle,  given  the 
hypotenuse  AB,  and  the  altitude. 

Fold  EF  (Fig.  21)  parallel  to  AB  at  the  distance 
of  the  given  altitude. 

Take  G  the  middle  point  of  AB.  Find  Hby  fold 
ing  GB  through  G  so  that  B  may  fall  on  EF. 


24 


GEOMETRIC  EXERCISES 


Fold  through  H  and  A,  G,  and  B. 
AHB  is  the  triangle  required. 


Fig.  21. 

63.  ABCD  (Fig.  22)  is  a  rectangle.    It  is  required 
to  find  a  square  equal  to  it  in  area. 

Q P 


R  O  B 

Fig.  22. 


M 


Mark  Q 

Find  <9,  the  middle  point  of  AM,  by  folding. 


IN  PAPER  FOLDING  25 

Fold  OM,  keeping  O  fixed  and  letting  M  fall  on 
line  BC,  thus  finding  P,  the  vertex  of  the  right-angled 
triangle  AMP. 

Describe  on  PB  the  square  BPQR. 

The  square  is  equal  to  the  given  rectangle. 

For  •.  •  BP  =  QP,  and  the  angles  are  equal,  triangle 
BMP  is  evidently  congruent  to  triangle  QSP. 

.-.  QS=BM=AD. 

.  - .  triangles  DA  T  and  QSP  are  congruent. 

.-.  PC=SR  and  triangles  RSA  and  CPT  are  con 
gruent. 

.-.  rnABCD  can  be  cut  into  three  parts  which 
can  be  fitted  together  to  form  the  square  RBPQ. 


Fig.  23. 

64.  Take  four  equal  squares  and  cut  each  of  them 
into  two  pieces  through  the  middle  point  of  one  of  the 
sides  and  an  opposite  corner.  Take  also  another 


26 


GEOMETRIC  EXERCISES 


equal  square.  The  eight  pieces  can  be  arranged  round 
the  square  so  as  to  form  a^complete  square,  as  in  Fig. 
23,  the  arrangement  being  a  very  interesting  puzzle. 

The  fifth  square  may  evidently  be  cut  like  the 
others,  thus  complicating  the  puzzle. 

65.  Similar  puzzles  can  be  made  by  cutting  the 
squares  through  one  corner  and  the  trisection  points 
of  the  opposite  side,  as  in  Fig.  24. 


Fig.  24. 

66.  If  the  nearer  point  is  taken  10  squares  are  re 
quired,  as  in  Fig.  24;  if  the  remoter  point  is  taken  13 
squares  are  required,  as  in  Fig.  25. 

67.  The  puzzles  mentioned  in  §§  65,  66,  are  based 
upon  the  formulas 


12  -I-  32  =  10 

22-f  32  =  13. 


IN  PAPER  FOLDING 


27 


The  process  may  be  continued,  but  the  number  of 
squares  will  become  inconveniently  large. 

68.  Consider  again  Fig.  13  in  §  44.  If  the  four 
triangles  at  the  corners  of  the  given  square  are  re 
moved,  one  square  is  left.  If  the  two  rectangles  FK 
and  KG  are  removed,  two  squares  in  juxtaposition 
are  left. 


Pi*.  25. 

69.  The  given  square  may  be  cut  into  pieces  which 
can  be  arranged  into  two  squares.  There  are  various 
ways  of  doing  this.  Fig.  23,  in  §  65,  suggests  the 
following  elegant  method :  The  required  pieces  are 
(1)  the  square  in  the  center,  and  (2)  the  four  con 
gruent  symmetric  quadrilaterals  at  the  corners,  to 
gether  with  the  four  triangles.  In  this  figure  the  lines 
from  the  mid-points  of  the  sides  pass  through  the  cor- 


28 


GEOMETRIC  EXERCISES 


ners  of  the  given  square,  and  the  central  square  is 
one  fifth  of  it.  The  magnitude  of  the  inner  square 
can  be  varied  by  taking  other  points  on  the  sides  in 
stead  of  the  corners. 

70.  The  given  square  can  be  divided  as  follows 
(Fig.  26)  into  three  equal  squares  : 

Take  £G  =  hali  the  diagonal  of  the  square. 


Fig.  26. 

Fold  through  C  and  G. 
Fold  BM  perpendicular  to  CG. 
Take  MP,  CN,  and  NL  each  =  BM. 
Fold  PH,  NK,  LF  at  right  angles  to  CG,  as  in 
Fig.  26. 


IN  PAPER  FOLDING  29 

Take  NK=BM,  and  fold  KE  at  right  angles  to 
NK. 

Then  the  pieces  1,  4,  and  6,  3  and  5,  and  2  and  7 
form  three  equal  squares. 

Now  CGi  =  ?>BGi, 
and  from  the  triangles  GBC  and  CMB 

BM  _BG 
~BC~~~CG} 
Letting  BC=a,  we  have 


- 
1/3 


IV.  THE  PENTAGON. 


71.  To  cat  off  a  regular  pentagon  from  the  square 
ABCD. 

Divide  BA  in  X  in  median  section  and  take  M 
the  mid-point  of  AX. 

D c 


A  M  X  N  S 

Fig.  27. 

Then  AB-AX=XB'1,  and  AM=MX. 
Take  BN=AM  or  MX. 


Lay  off  NP  and  J/^  equal  to  MN,  so  that  /*  and 
R  may  lie  on  BC  and  ^4Z>  respectively. 


PAPER  FOLDING  3: 


Lay  off  RQ  and  PQ  =  MR  and  NP. 

MNPQR  is  the  pentagon  required. 

In  Fig.  19,  p.  22,  AN,  which  is  equal  to  AB,  has 
the  point  JV  on  the  perpendicular  MO.  If  A  be  moved 
on  ^4.Z?  over  the  distance  MB  ',  then  it  is  evident  that 
.A7'  will  be  moved  on  to  BC,  and  J^f  to  Jf. 

Therefore,  in  Fig.  27,  NR  =  AB.  Similarly  MP  = 
Ab  RP  is  also  equal  to  AB  and  parallel  to  it. 

/  RMA  =  |  of  a  rt.  /  . 
.  •  .  /  NMR  =  f  of  a  rt  /  . 

Similarly        /  PNM=  f  of  a  rt.  /  . 

From  triangles  MNR  and  QRP,  /  NMR=  £RQP 
=  £  of  art.  /. 

The  three  angles  at  J/,  TV,  and  Q  of  the  pentagon 
being  each  equal  to  £  of  a  right  angle,  the  remain 
ing  two  angles  are  together  equal  to  -^2-  of  a  right 
angle,  and  they  are  equal.  Therefore  each  of  them 
is  |  of  a  right  angle. 

Therefore  all  the  angles  of  the  pentagon  are  equal. 

The  pentagon  is  also  equilateral  by  construction. 

72.  The  base  MN  of  the  pentagon  is  equal  to  XB, 
\   e.,  to^?-  (l/5  —  1)=^X0.6180....  §  58. 

The  greatest  breadth  of  the  pentagon  is  AB. 

73.  If/  be  the  altitude, 


GEOMETRIC  EXERCISES 


8 


215 


0.9510. . . .  =AB  cos 


Fig.  28. 

74.  if  J?  be  the  radius  of  the  circumscribed  circ)e, 

x=    AB  =     ZAS 

2cosl8°       T/io  +  21/5 


o 


IN  PAPER  FOLDING  33 

75.  If  r  be  the  radius  of  the  inscribed  circle,  then 
from  Fig.  28  it  is  evident  that 


20 


0.4253.... 

76.  The  area  of  the  pentagon  is  5r  X  i  the  base  of 
the  pentagon,  i.  e., 


10 


77.  In  Fig.  27  let  PR  be  divided  by  MQ  and 
in  .£  and  F. 


Then  -.-  MN=  ±£l.(i/5  — 1)  ...  §  72 

a 

and  cos  36°=   -   --^-l)' 


cos  36 


5  _|_ 


34  GEOMETRIC  EXERCISES 


-2)...  (2) 
RF=MN. 

RF:  RE  =  RE  :  £F(by  §  51)  ....................  (3) 

1/5  —  1:3  —  1/5  =  3  —  1/5  :  2  (1/5  —  2)  ...........  (4) 

By  §  76  the  area  of  the  pentagon 


i  •  1/25  +  101/5, 


snce          =  V  5 


2 
.  •.  the  area  of  the  inner  pentagon 

=  EF*  -  l  •  1/25+  101/5 


1 


=  ^^2  -(1/5  —  2)2  •  ~.-  I/  25  +  101    5  . 

The  larger  pentagon  divided  by  the  smaller 


=  2  :  (7  —  3l/5) 
=  1  :  0.145898..  .. 


78.  If  in  Fig.  27,  angles  (^^  and  LFQ  are  made 
equal  to  ERQ  or  FQP,  K,  L  being  points  on  the  sides 
QR  and  QP  respectively,  then  EFL  QK  will  be  a  reg 
ular  pentagon  congruent  to  the  inner  pentagon.  Pen 
tagons  can  be  similarly  described  on  the  remaining 
sides  of  the  inner  pentagon.  The  resulting  figure 
consisting  of  six  pentagons  is  very  interesting. 


V.   THE  HEXAGON. 

79.  To   cut  off   a   regular   hexagon   from   a   given 
square. 


Fig.  29. 


Fold  through  the  mid-points  of  the  opposite  sides, 
and  obtain  the  lines  A  OB  and  COD. 

On  both  sides  of  AO  and  OB  describe  equilateral 
triangles  (§  25),  AOE,  AHO;  BFO  and  BOG. 


36  GE  OME  TRIG  EXER  CISES 

Draw  EF  and  HG. 

AHGBFE  is  a  regular  hexagon. 

It  is  unnecessary  to  give  the  proof. 

The  greatest  breadth  of  the  hexagon  is  AB. 

80.  The  altitude  of  the  hexagon  is 
=  0.866.... 


Fig.  30. 

81.  If  R  be  the  radius  of  the  circumscribed  circle, 

R=\AB. 

82.  If  r  be  the  radius  of  the  inscribed  circle, 


r  =        -  '  A  B  =  °  •  4  3  3  •  •  •  •  X  A  B. 
4 


IN  PAPER  FOLDING  37 

83.  The  area  of  the  hexagon  is  6  times  the  area  of 
the  triangle  HGO, 

,  =  6.^-1^. 

=  3T/  3   -.4^2^0.6495 X^^2- 

8 

Also  the  hexagon  =  J  •  -42?  •  CD. 

=  14  times  the  equilateral  triangle  on  AB. 


Fig.  31- 

84.  Fig.  -30  is  an  example  of  ornamental  folding 
into  equilateral  triangles  and  hexagons. 

85.  A  hexagon  is  formed  from  an  equilateral  tri 
angle  by  folding  the  three  corners  to  the  center. 

The  side  of  the  hexagon   is  i  of  the  side  of  the 
equilateral  triangle. 


38  GEOMETRIC  EXERCISES 

The  area  of  the  hexagon  =  f  of  the  equilateral 
triangle. 

86.  The  hexagon  can  be  divided  into  equal  regular 
hexagons  and  equilateral  triangles  as  in  Fig.  31  by 
folding  through  the  points  of  trisection  of  the  sides. 


VI.   THE  OCTAGON. 

87.  To  cut  off  a  regular  octagon  from  a  given  square. 
Obtain  the  inscribed  square  by  joining  the  mid 
points  A,  B,  C,  D  of  the  sides  of  the  given  square. 


Fig.  32. 

Bisect  the  angles  which  the  sides  of  the  inscribed 
square  make  with  the  sides  of  the  other.  Let  the  bi 
secting  lines  meet  in  E,  F,  G,  and  H. 


40  GEOMETRIC  EXERCISES 

AEBFCGDH'is  a  regular  octagon. 

The  triangles  AEB,  BFC,  CGD,  and  DHA  are 
congruent  isosceles  triangles.  The  octagon  is  there 
fore  equilateral. 

The  angles  at  the  vertices,  E,  F,  G,  H  of  the  same 
four  triangles  are  each  one  right  angle  and  a  half, 
since  the  angles  at  the  base  are  each  one-fourth  of  a 
right  angle. 

Therefore  the  angles  of  the  octagon  at  A,  B,  C, 
and  D  are  each  one  right  angle  and  a  half. 

Thus  the  octagon  is  equiangular. 

The  greatest  breadth  of  the  octagon  is  the  side  of 
the  given  square,  a. 

88.  If  R  be  the  radius  of  the  circumscribed  circle, 
and  a  be  the  side  of  the  original  square, 

*-f 

89.  The  angle  subtended  at  the  center  by  each  of 
the  sides  is  half  a  right  angle. 

90.  Draw  the  radius  OE  and  let  it  cut  AB  in  K 
(Fig.  33). 


Then  AK=  OK=          =  -=r. 
V  2       2  V  2 

KE=OA-OK=a- 


2]   2 

Now  from  triangle  AEK, 


IN  PAPER  FOLDING 


41 


=  4--  (4-21/2) 


=  -£-.(2-1/2). 


=-  1/2-1/2. 


91.  The  altitude  of  the  octagon  is  CE  (Fig.  33). 
But  C£2  =  AC2  — 


=  *»—-•  (2  _  1/2  )  =        •  (2  +  1/2) 


A 

Fig.  33- 


a       / 7^" 

2 

92.  The  area  of  the  octagon  is  eight  times  the  tri 
angle  AOE  and 

o 

CL  d  d 


42 


GEOMETRIC  EXERCISES 


93.  A  regular  octagon  may  also  be  obtained  by 
dividing  the  angles  of  the  given  square  into  four  equal 
parts. 

2  _  _  Y 


Fig.  34- 

It  is  easily  seen  that  EZ=  WZ  =  a,  the  side  of  the 
square. 


=  WK; 


XE 


•.•(2—1/1). 


Now 
.-. 

Also 


a±  —  21/2. 
XZ—(1XE 


a.tf(2—  1/2). 


IN  PAPER  FOLDING 


Again  OZ= 


=  ~(6-4v/2  +  2) 
=  <z2(2  — 1/2"). 


HK=KZ—HZ 


=  a-  (1/2  —  1/2  )  -(1/2  —  1) 


—  7i/2 


and  HA  =  ~  1/20  —  141/2. 
A 

94.  The  area  of  the   octagon  is   eight  times   the 
area  of  the  triangle  HOA, 


1/2 


(6  —  4i/2) 

2— 4) 
=  «2- 1/2 -(1/2— -I)2. 


44  GEOMETRIC  EXERCISES 

95.  This  octagon  :  the  octagon  in  §  92 

=  (2  — 1/2  )2  : 1  or  2  :  (j/2  +  I)2; 
and  their  bases  are  to  one  another  as 
1/2":  i/2"+  1. 


VII.   THE  NONAGON. 

96.  Any  angle  can  be  trisected  fairly  accurately  by 
paper  folding,  and  in  this  way  we  may  construct  ap 
proximately  the  regular  nonagon. 


Fig.  35- 


Obtain  the  three  equal  angles  at  the  center  of  an 
equilateral  triangle.      (§  25.) 


46  GEOMETRIC  EXERCISES 

For  convenience  of  folding,  cut  out  the  three 
angles,  A  OF,  FOC,  and  CO  A. 

Trisect  each  of  the  angles  as  in  Fig.  35,  and  make 
each  of  the  arms  =  OA. 

97.  Each  of  the  angles  of  a  nonagon  is  -^4-  of  a 
right  angle  =  140°. 

The  angle  subtended  by  each  side  at  the  center  is 
J  of  a  right  angle  or  40°. 

Half  this  angle  is  \  of  the  angle  of  the  nonagon. 

98.  OA  =  %a,  where  a  is  the  side  of  the  square  ;  it 
is  also  the  radius  of  the  circumscribed  circle,  R. 

The  radius  of  the  inscribed  circle  =Jt  .  cos  20° 
=  %a  cos  20° 

=  *  X0.  9396926 


=  aX  0.4698463. 

The  area  of  the  nonagon  is  9  times  the  area  of 
the  triangle  AOL 


=  f  7?2-  sin  40° 


, 
=  --X  0.6427876 

o 

=  a*  X  0.723136. 


VIII.  THE  DECAGON  AND  THE  DODECAGON 

99.  Figs.  36,  37  show  how  a  regular  decagon,  and 
a  regular  dodecagon,  may  be  obtained  from  a  penta 
gon  and  hexagon  respectively. 


Fig.  36. 

The  main  part  of  the  process  is  to  obtain  the 
angles  at  the  center. 

In  Fig.  36,  the  radius  of  the  inscribed  circle  of 
the  pentagon  is  taken  for  the  radius  of  the  circum 
scribed  circle  of  the  decagon,  in  order  to  keep  it 
within  the  square. 


48  GEOMETRIC  EXERCISES 

ioo.  A  regular  decagon  may  also  be  obtained  'as 
follows  : 

Obtain  X,  Y,  (Fig.  38),  as  in  §  51,  dividing  AB  in 
median  section. 

Take  M  the  mid-point  of  AB. 

Fold  XC,  MO,  YD  at  right  angles  to  AB. 

Take  O  in  MO  such  that  YO  =  AY,  or  YO  =  XB. 


Fig.  37- 

Let  YO,  and  XO  produced  meet  XC,  and  YD  in  C 
and  D  respectively. 

Divide  the  angles  XOC  and  DOY  into  four  equal 
parts  by  HOE,  KOF,  and  LOG. 

Take  Off",  OK,  OL,  OE,  OF,  and  OG  equal  to 
<9For  OK 

Join  X,  H,  K,  L,  C,  D,  E,  F,  G,  and  Y,  in  order. 


IN  PAPER  FOLDING 

As  in  §  60, 

/  YOX=l  of  a  rt.  /  =36°. 


49 


Fig.  38. 

By  bisecting  the  sides  and  joining  the  points  thus 
determined  with  the  center,  the  perigon  is  divided 
into  sixteen  equal  parts.  A  16-gon  is  therefore  easily 
constructed,  and  so  for  a  32-gon,  and  in  general  a. 
regular  2*-gon. 


IX.  THE  PENTEDECAGON. 


101.  Fig.   39  shows  how  the  pentedecagon  is  OD- 
tained  from  the  pentagon. 

Let  ABODE  be  the  pentagon  and  O  its  center 


Draw  OA,  OB,  OC,  OD,  and  OE.  Produce  DO 
to  meet  AB  in  K. 

Take  OF=$  of  OD. 

Fold  GFH  2,\.  right  angles  to  OF.  Make  OG  = 
OH=  OD. 


PAPER  FOLDING  51 

Then    GDH.  is   an   equilateral    triangle,    and    the 
angles  DOG  and  HOD  are  each  120°. 

But  angle  DOA  is  144°;  therefore  angle  GOA  is 
24°. 

That  is,  the  angle  EOA,  which  is  72°,  is  trisected 
by  OG. 

Bisect  the  angle  EOG  by  OL,  meeting  EA  in  Z, 
and  let  OG  cut  EA  in  M;  then 
OL  =  OM. 

In  OA  and  <9^  take  OP  and  <9<2  equal  to  OL  or 


Then  PM,  ML,  and  Z<2  are  three  sides  of  the 
pentedecagon. 

Treating  similarly  the  angles  A  OB,  BOC,  COD, 
and  DOE,  we  obtain  the  remaining  sides  of  the  pente 
decagon. 


X.   SERIES. 


ARITHMETIC  SERIES. 


IO2.  Fig.  40  illustrates  an  arithmetic  series.  The 
horizontal  lines  to  the  left  of  the  diagonal,  including 
the  upper  and  lower  edges,  form  an  arithmetic  series. 


Fig.  40. 

The  initial  line  being  a,  and  d  the  common  difference, 
the  series  is  a,  a-\-  d,  a-\-2d,  a -\-  3//,  etc. 

103.  The  portions  of  the  horizontal  lines  to   the 
right  of  the  diagonal  also  form   an  arithmetic  series, 


PAPER  FOLDING  53 

but  they  are   in   reverse  order  and  decrease  with   a 
common  difference. 

104.  In  general,  if  /  be  the  last  term,  and  s  the 
sum  of  the  series,  the  above  diagram  graphically 
proves  the  formula 


105.  If  a  and  c  are  two  alternate  terms,  the  middle 
term  is 


106.  To  insert  n  means  between  a  and  /,  the  ver 
tical  line  has  to  be  folded  into  n-\-  1  equal  parts.  The 
common  difference  will  be 

I—-  a 


107.  Considering  the  reverse  series  and  interchan 
ging  a  and  /,  the  series  becomes 

a,  a  —  d,  a  —  2d '. .  . .  /. 

The  terms  will  "be  positive  so  long  as  a  >(» — !)</, 
and  thereafter  they  will  be  zero  or  negative. 

GEOMETRIC  SERIES. 

108.  In  a  right-angled  triangle,  the  perpendicular 
from  the  vertex  on  the  hypotenuse  is  a  geometric  mean 
between  the  segments  of  the  hypotenuse.      Hence,  if 
two   alternate   or   consecutive   terms    of    a   geometric 
series  are  given  in  length,   the   series   can    be  deter- 


54 


GEOMETRIC  EXERCISES 


mined   as  in   Fig.   41.      Here   OPi,    OP2,    OP3, 

and  OP$  form  a  geometric  series,  the  common  rate 

being  OPi  :  OP2. 

P3 


Fig.  41. 


If  OP\  be  the  unit  of  length,  the  series  consists  of 
the  natural  powers  of  the  common  rate. 

109.   Representing  the  series  by  #,  ar,  ar2, .... 


These  lines  also  form  a  geometric  series  with  the 
common  rate  r. 

no.  The  terms  can  also  be  reversed,  in  which  case 
the  common  rate  will  be  a  proper  fraction.  If  OP& 
be  the  unit,  OP±  is  the  common  rate.  The  sum  of 
the  series  to  infinity  is 


IN  PAPER  FOLDING  55 

in.  In  the  manner  described  in  §  108,  one  geo 
metric  mean  can  be  found  between  two  given  lines, 
and  by  continuing  the  process,  3,  7,  15,  etc.,  means 
can  be  found.  In  general,  2"  —  1  means  can  be  found, 
n  being  any  positive  integer. 

112.  It  is  not  possible  to  find  two  geometric  means 
between  two  given  lines,  merely  by  folding  through 
known  points.      It  can,  however,  be  accomplished   in 
the  following  manner  :  In  Fig.  41,  OP\  and  OP±  being 
given,  it  is  required  to  find  P%  and  Ps.    Take  two  rect 
angular  pieces  of  paper  and  so  arrange  them,  that  their 
outer  edges  pass  through  P\  and  P^  and  two  corners 
lie  on  the  straight  lines  OP*  and  OP*  in  such  a  way 
that  the  other  edges  ending  in  those  corners  coincide. 
The  positions  of  the  corners  determine  OP<i  and  OP*. 

113.  This  process  gives  the  cube  root  of  a  given 
number,  for  if  OP\  is  the  unit,  the  series  is  1,  r,  r2,  r3. 

114.  There  is  a  very  interesting  legend  in  connec 
tion  with  this  problem.*      "The  Athenians  when  suf 
fering  from  the  great  plague  of  eruptive  typhoid  fever 
in  430  B.  C.,  consulted  the  oracle  at  Delos  as  to  how 
they  could   stop    it.      Apollo   replied  that    they  must 
double  the  size  of  his  altar  which  was  in  the  form  of  a 
cube.      Nothing  seemed  more  easy,   and   a   new  altar 
was  constructed  having  each  of  its  edges  double  that 
of  the  old  one.     The  god,  not  unnaturally  indignant, 

*But  see  Beman  and  Smith's  translation  of  Fink's  History  of  Mathe 
matics,  p.  82,  207. 


56  GEOMETRIC  EXERCISES 

made  the  pestilence  worse  than  before.  A  fresh  dep 
utation  was  accordingly  sent  to  Delos,  whom  he  in 
formed  that  it  was  useless  to  trifle  with  him,  as  he 
must  have  his  altar  exactly  doubled.  Suspecting  a 
mystery,  they  applied  to  the  geometricians.  Plato, 
the  most  illustrious  of  them,  declined  the  task,  but 
referred  them  to  Euclid,  who  had  made  a  special 
study  of  the  problem."  (Euclid's  name  is  an  inter 
polation  for  that  of  Hippocrates.)  Hippocrates  re 
duced  the  question  to  that  of  finding  two  geometric 
means  between  two  straight  lines,  one  of  which  is 
twice  as  long  as  the  other.  If  a,  x,  y  and  2a  be  the 
terms  of  the  series,  xs  =  2a^.  He  did  not,  however, 
succeed  in  rinding  the  means.  Menaechmus,  a  pupil 
of  PJato,  who  lived  between  375  and  325  B.  C.,  gave 
the  following  three  equations  :  * 

a  :  x  =  x  :  y  —y  :  2a. 

From  this  relation  we  obtain  the  following  three 
equations  : 

x2  =  ay    (1) 

y2  =  2ax (2) 

xy  =  2a* (3) 

(1)  and  (2)  are  equations  of  parabolas  and  (3)  is 
the  equation  of  a  rectangular  hyperbola.  Equations 
(1)  and  (2)  as  well  as  (1)  and  (3)  give  x*  =  2a*.  The 
problem  was  solved  by  taking  the  intersection  («)  of 
the  two  parabolas  (1)  and  (2),  and  the  intersection  (/?) 
of  the  parabola  (1)  with  the  rectangular  hyperbola  (3). 

*Ibid.,  p.  207. 


IN  PAPER  FOLDING  57 


HARMONIC  SERIES. 

115.  Fold  any  lines    AR,  PB,  as  in  Fig.  42,  P  be 
ing  on  AR,  and  B  on  the  edge  of  the  paper.      Fold 
again  so  that  AP  and  PR  may  both  coincide  with  PB. 
Let  PX,  PY  be  the  creases  thus  obtained,  X  and  Y 
being  on  AB. 

Then  the  points  A,  X,  B,  Y  form  an  harmonic 
range.  That  is,  AB  is  divided  internally  in  X  and 
externally  in  Fso  that 

AX:  XB  =  AY:  BY. 

It  is  evident,  that  every  line  cutting  PA,  PX,  PB, 
and  PYvti\\  be  divided  harmonically. 

R 


A  X  B  Y 

Fig.  42. 

116.  Having  given  A,  B,  and  X,  to  find  Y:  fold  any 
line  XPand  mark  ^corresponding  to  B.  Fold  AKPR, 
and  BP.      Bisect  the  angle  BPR  by  PY  by  folding 
through  P  so  that  /l#  and  PR  coincide. 
Because  XP  bisects  the  angle  APB, 
.-.AX:  XB  =  AP:  BP, 
=  AY:  BY. 


58  GEOMETRIC  EXERCISES 

117.  AX:  XB  =  AY:  BY 
orAY—XY:  XY—BY=AY:  BY. 

Thus,  AY,  XY,  and  BY,  are  an  harmonic  series, 
and  XY  is  the  harmonic  mean  between  AY  and  BY. 

Similarly  AB  is  the  harmonic  mean  between  AX 
and  A  Y. 

118.  If  BY  and  XYbe  given,  to  find  the  third  term 
AY,   we  have  only  to  describe  any  right-angled   tri 
angle  on  XYas  the  hypotenuse  and  make  angle  APX 
=  angle  XPB. 

119.  Let  AX— a,  AB  =  b,  and  AY=c. 


a  -j-  c 
or,  ab  -\-  bc=.%ac 

ab               b 
or,  ^  — ~o 7  = T- 

a 

When  a  =  fr,  c  =  b. 
When  b  =  2a,  c=&. 

Therefore  when  X  is  the  middle  point  of  AB,  Y  is 
at  an  infinite  distance  to  the  right  of  B.  Y  approaches 
B  as  X  approaches  it,  and  ultimately  the  three  points 
coincide. 

As  X  moves  from  the  middle  of  AB  to  the  left,  Y 
moves  from  an  infinite  distance  on  the  left  towards  Ay 
and  ultimately  X,  A,  and  Y  coincide. 

120.   If  E  be  the  middle  point  of  AB, 


for  all  positions  of  Jfand  Y  with  reference  to  A  or  B. 


IN  PAPER  FOLDING  59 

Each  of  the  two  systems  of  pairs  of  points  X  and 
Y  is  called  a  system  in  involution,  the  point  E  being 
called  the  center  and  A  or  B  the  focus  of  the  system. 
The  two  systems  together  may  be  regarded  as  one 
system. 

121.  AX  and  AY  being  given,  B  can  be  found  as 
follows : 

Produce  XA  and  take  AC=XA. 

Take  D  the  middle  point  of  A  Y. 

Take  CE  =  DA  or  AE  =  DC. 

F 


£  C          A        X       B  D  Y 

Fig.  43- 

Fold  through  A  so  that  AF  may  be  at  right  angles 
to  CA  Y. 

Find  F  such  that  DF=DC. 

Fold  through  EF  and  obtain  FB,  such  that  FB  is 
at  right  angles  to  EF. 

CD  is  the  arithmetic  mean  between  AX  and  A  Y. 

AF  is  the  geometric  mean  between  AX  and  A  Y. 

AFis  also  the  geometric  mean  between  CD  or  AE 
and  AB. 

Therefore  AB  is  the  harmonic  mean  between  AX 
and  A  Y. 

122.  The  following  is  a  very  simple  method  of 
finding  the  harmonic  mean  between  two  given  lines. 


6o 


GEOMETRIC  EXERCISES 


Take  AB,  CD  on  the  edges  of  the  square  equal  to 
the  given  lines.  Fold  the  diagonals  AD,  ^Cand  the 
sides  AC,  BD  of  the  trapezoid  ACDB.  Fold  through 
E,  the  point  of  intersection  of  the  diagonals,  so  that 
PEG  may  be  at  right  angles  to  the  other  sides  of  the 
square  or  parallel  to  AB  and  CD.  Let  EEG  cut  AC 
A  B 


For 


Fig.  44. 

and  BD  in  F  and  G.    Then  EG  is  the  harmonic  mean 
between  AB  and  CD. 

EE  __  CE 
AB  ~~  CB 
EG  __  FE  __  EB_ 
CD  ~~  CD  ~~  ~C~B 

CE        EB_ 
CB  H"  CB  ~ 

1  1  1          _2_ 

~AB  +  CD  ~=  ~FE  ~~~  ~EG' 


EE  _    EF 
AB    '    CD 


IN  PAPER  FOLDING 


6t 


123.  The  line  HK  connecting  the  mid-points  of  AC 
and  BD  is  the  arithmetic  mean  between  AB  and  CD. 

124.  To  find  the  geometric  mean,  take  HL  in  HK 
=  FG.    Fold  ZJ/at  right  angles  to  HK.    Take  O  the 
mid-point  of  HK  and  find  M  in  LM  so  that  OM^OH. 
HM\s  the  geometric  mean  between  ^^  and  CD  as  well 
as  between  FG  and  ZfA".   The  geometric  mean  between 
two  quantities  is  thus  seen  to  be  the  geometric  mean 
between  their  arithmetic  mean  and  harmonic  mean. 


O 

A 

B 

C 

D 

E 

F 

a 

b 

c 

a 

e 

f 

Fig.  45- 
SUMMATION  OF  CERTAIN  SERIES. 

125.  To  sum  the  series 

1  +  3  +  5....  +  (2»  — 1). 

Divide  the  given  square  into  a  number  of  equal 
squares  as  in  Fig.  45.  Here  we  have  49  squares,  but 
the  number  may  be  increased  as  we  please. 


62 


GEOMETRIC  EXERCISES 


The  number  of  squares  will  evidently  be  a  square 
number,  the  square  of  the  number  of  divisions  of  the 
sides  of  the  given  square. 

Let  each  of  the  small  squares  be  considered  as  the 
unit;  the  figure  formed  \yy  A -\- O -\- a  being  called  a 
gnomon. 

The  numbers  of  unit  squares  in  each  of  the  gno 
mons  AOa,  BOb,  etc.,  are  respectively  3,  5,  7,  9, 
11,  13. 

Therefore  the  sum  of  the  series  1,  3,  5,  7,  9,  11, 
13  is  72. 

Generally,  1  +  3  +  5  -f  . .  . .  -f  (2»  —  1)  =  ;/2. 


o 

A 

B 

c 

D 

E 

r 

1 

Z 

3 

4 

5 

b 

7 

a 

Z 

4 

fe 

g 

IO 

IZ 

14 

lo 

3 

b 

q 

IZ 

IS 

18 

Zl 

c 

4 

8 

IZ 

|(o 

zo 

Z4 

Z? 

d 

5 

10 

1  S 

ZO 

Z5 

30 

35 

e 

fc 

12. 

u 

Z4 

30 

3k 

42 

f 
7 

14 

Zl 

2? 

35 

42 

4<? 

Fig.  46. 

126.  To  find   the  sum  of  the  cubes  of  the  first  n 
natural  numbers. 

Fold  the  square  into  49  equal  squares  as  in  the 


IN  PAPER  FOLDING  63 

preceding  article,  and  letter  the  gnomons.   Fill  up  the 
squares  with  numbers  as  in  the  multiplication  table. 

The  number  in  the  initial  square  is  1  =  I3. 

The  sums  of  the  numbers  in  the  gnomons  Aa,  Bb, 
etc.,  are  2  +  4-f  2  =  23,  33,  43,  53,  63,  and  73. 

The  sum  of  the  numbers  in  the  first  horizontal 
row  is  the  sum  of  the  first  seven  natural  numbers. 
Let  us  call  it  s. 

Then  the  sums  of  the  numbers  in  rows  a,  b,  c,  d, 
etc.,  are 

2s,  3s,  4s,  5j,  6s,  and  Is. 

Therefore  the  sum  of  all  the  numbers  is 
s(l  +  2  +  3  +  4  +  5  +  6  +  7)  =  s*. 

Therefore,  the  sum  of  the  cubes  of  the  first  seven 
natural  numbers  is  equal  to  the  square  of  the  sum  of 
those  numbers. 

Generally,  I3  +  23  -f  33  ____  -f  ;z3 


For     [«•(«  +  I)]2—  [(»—  !)•«]* 

=--  (>2  -f  «)«  —  (>2  —  «)2  = 

Putting  «  =  1,  2,  3  ....  in  order,  we  have 
4-l3  =  (l-2)2_  (0-1)2 


64  GE  OME  TRIG  EXER  CISES 

Adding  we  have 

4.51/*3  =  [«(»  + I)]2 


2 
127-  If  •*•„  be  the  sum  of  the  first  n  natural  numbers, 

128.  To  sum  the  series 

1-2  +  2-3  +  3-4. ...  +  («—  !)-». 

In  Fig.  46,  the  numbers  in  the  diagonal  commen 
cing  from  1,  are  the  squares  of  the  natural  numbers 
in  order. 

The  numbers  in  one  gnomon  can  be  subtracted 
from  the  corresponding  numbers  in  the  succeeding 
gnomon.  By  this  process  we  obtain 

+  2[«(«  —  !)  +  («  —  2)  +  («  — 3)....  +1] 
=«*  +  («— 1)2 +  .2[;  +  2..  ..+(*—!)] 

—  1  4.  3(n  —  l)n. 
Now   •.•  «3  — (;?— 1)3  =  1 -f- 3(« — 1)», 


93  —  p^!          3.  9-1 


Hence,  by  addition, 


IN  PAPER  FOLDING  65 

Therefore 


129.  To  find  the  sum  of  the  squares  of  the  first  n 
natural  numbers. 

1-2  +  2-3..  ..  +  («_!)•» 

=  22  —  2  +  32  —  3..  ..  +  ^2  —  a 

=  12  _|_  22  +  32  ____  +  rc2  —  (1  4-  2  -f  3  ____  -f  n) 


Therefore 


130.  To  sum  the  series 

12_|_32_|_  52 ^  (2«  —  I)2. 

«8_c«_i)s==:W2+  („_!)»  +  »(«—.!),  by  §  128, 

=  (2»  — 1)2-(^  —  !)-», 
.  • .   by  putting  n  =  1,  2,  3, .... 
13_o3=zi2__o.l 
23  — Pr=32  — 1-2 
33  — 23  =  52  — 2-3 


66  GEOMETRIC  EXERCISES 

Adding,  we  have 

n*  =  I2  4.  32  +  52 . .  . .  -{-  (2«  —  I)2 

—  [1-2  -f  2-34-3-4..  ..  +  (»_  !)•«], 


XI.   POLYGONS. 

I3lt  Find  O  the  center  of  a  square  by  folding  its 
diameters.  Bisect  the  right  angles  at  the  center, 
then  the  half  right  angles,  and  so  on.  Then  we 
obtain  2"  equal  angles  around  the  center,  and  the 
magnitude  of  each  of  the  angles  is  ^  of  a  right  angle, 
«  being  a  positive  integer.  Mark  off  equal  lengths  on 
each  of  the  lines  which  radiate  from  the  center.  If 
the  extremities  of  the  radii  are  joined  successively, 
we  get  regular  polygons  of  2"  sides. 

132.  Let  us  find  the  perimeters  and  areas  of  these 
polygons.  In  Fig.  47  let  OA  and  OA\  be  two  radii 
at  right  angles  to  each  other.  Let  the  radii  OA^ 
OAS,  OA4,  etc.,  divide  the  right  angle  A\OA  into  2, 
4,  8 ....  parts.  Draw  AA\,  AA-2,  AA3 ....  cutting  the 

radii  OA-2,  OA^  OA± at  B\,  B-2,  B3 respectively, 

at  right  angles.  Then  B\,  B^  £3 . .  . .  are  the  mid 
points  of  the  respective  chords.  Then  AA\,  AA<t, 
A AB,  AA^, .  .  .are  the  sides  of  the  inscribed  polygons 

of  22,  23,  24 sides  respectively,  and  OB^  OB* 

are  the  respective  apothems. 

Let  OA  =  R, 

a  (2")  represent  the  side  of  the  inscribed  polygon 


68  GEOMETRIC  EXERCISES 

of  2"  sides,  £(2")  the  corresponding  apothem,  /(2")  its 
perimeter,  and  .4(2")  its  area. 
For  the  square, 

/  (22)=  ^-22- 1/2; 


Fig.  47- 


For  the  octagon, 
in  the  two  triangles 


and 
OA 


A, 


or  AAv  = 


(1) 


PAPER  FOLDING 


2  —  1/2 


6g 

(2) 


— 


=  1*1/2+1/2    ...(3) 


23)  =  J  perimeter  X  apothem 


—  1/2  - 


Similarly  for  the  polygon  of  16  sides, 


2*)  =  *-24 -1/2  — 1/2  +  1/2; 


^(24)  =  ^2-  22-1/2  — 1/2; 
and  tor  the  polygon  of  32  sides, 


25)  =  ^-25-  1/2 


The  general  law  is  thus  clear. 
Also  ^2«==- 


As  the  number  of  sides   is  increased  indefinitely 


7o 


GEOMETRIC  EXERCISES 


the  apothem  evidently  approaches  its  limit,  the  ra 
dius.      Thus  the  limit  of 


2  +  I/  2  +  1/2..  ..is  2; 


for  if  jc  represent  the  limit,  #  =  1/2 -J- #,  a  quadratic 
which  gives  x  =  2,  or  — 1;  the  latter  value  is,  of 
course,  inadmissible. 

133.  If  perpendiculars  are  drawn  to  the  radii  at 
their  extremities,  we  get  regular  polygons  circum 
scribing  the  circle  and  also  the  polygons  described  as 
in  the  preceding  article,  and  of  the  same  number  of 
sides. 

C_ F E__       _G  D 


In  Fig.  48,  let  AE  be  a  side  of  the  inscribed  poly 
gon  and  FG  a  side  of  the  circumscribed  polygon. 
Then  from  the  triangles  FIE  and  EIO, 
OE        FE 


.-.  FG  =  R 


AE 


IN  PAPER  FOLDING  71 

The  values  of  AE  and  Of  being  known  by  the 
previous  article,  FG  is  found  by  substitution. 

The  areas  of  the  two  polygons  are  to  one  another 
as  FG1  :  AE'1,  i.  e.,  as  &• :  Of*. 

134.  In  the  preceding  articles  it  has  been  shown 
how  regular  polygons  can   be  obtained  of  22,  23. . . .2* 
sides.   And  if  a  polygon  of  m  sides  be  given,  it  is  easy 
to  obtain  polygons  of  2"-m  sides. 

135.  In   Fig.  48,  AB  and  CD  are  respectively  the 
sides  of  the  inscribed   and  circumscribed  polygons  of 
n  sides.      Take  E  the  mid  point  of  CD  and  draw  AE, 
BE.     AE  and  BE  are  the  sides  of  the  inscribed  poly 
gon  of  2n  sides. 

Fold  AF,  BG  at  right  angles  to  A  C  and  BD,  meet 
ing  CD  in  Fand  G. 

Then  FG  is  a  side  of  the  circumscribed  polygon 
of  2n  sides. 

Draw  OF,  OG  and  OE. 

Let  />,  P  be  the  perimeters  of  the  inscribed  and 
circumscribed  polygons  respectively  of  n  sides,  and 
A,  B  their  areas,  and  /,  P'  the  perimeters  of  the  in 
scribed  and  circumscribed  polygons  respectively  of  2n 
sides,  and  A',  B'  their  areas. 

Then 
p  =  n-AB,  P=n-CD,  p'  =  2n-AE,  P'  =  2n-FG. 

Because   OF  bisects   /  COE,   and  AB  is  parallel 

to  CD, 

CJ^__CO__     CO  _     CD 

FE  ~~  ~OE  ~~  ~AO  "    ~ 


72  GEOMETRIC  EXERCISES 

CE  _CD+AB 
' ' '  ~EE  ~  AB  ' 
4n-C£  n-CD+n-AB 


EE  n-AJ3 

2P  _P+p 


or 


.'.  P'  = 

Again,  from  the  similar  triangles  EIE  and  AffE, 
ET^     _  EE 
~Aff~  AJE' 
or  A£2=.2. 


or  p'  =  V  P'p. 
Now, 


The  triangles  A  Off  and  AOE  are  of  the  same  alti 
tude,  AH, 

OH 


__ 

'   &AOE  =~  ~OE' 
Similarly, 

_OA 
~  ~OC' 


Again  because  AB  ||  CD, 

A  A  Off        A  AOE 


Now  to  find  B  '.     Because  the  triangles  COE  and 


IN  PAPER  FOLDING 


73 


FOE   have  the   same   altitude,    and    OF  bisects   the 
angle  EOC, 

&COE        CE        OC+OE 


&FOE~    FE 

and   OE  =  OA, 


OE 


,__     _ 
OA  ~~  Off  ~  &AOH  ' 


&COE        &AOE  + 


&AOH 

From  this  equation  we  easily  obtain  . — IT . 

y  B'  A      ' 


136.  Given  the  radius  R  and  apothem  r  of  a  reg 
ular  polygon,  to  find  the  radius  R  and  apothem  r'  of 
a  regular  polygon  of  the  same  perimeter  but  of  double 
the  number  of  sides. 

Let  AB  be  a  side  of  the  first  polygon,  O  its  center, 
OA  the  radius  of  the  circumscribed  circle,  and  OD 
the  apothem.  On  OD  produced  take  OC=OA  or 
OB.  Draw  AC,  BC.  Fold  OA'  and  OB'  perpen- 


74  GEOMETRIC  EXERCISES 

dicular  to  AC  and  BC  respectively,  thus  fixing  the 
points  A',  B'.  Draw  A'B'  cutting  OC  in  D'.  Then 
the  chord  A'  B'  is  h^f  of  AB,  and  the  angle  B'OA'  is 
half  of  BOA.  OA'  and  OH  are  respectively  the  ra 
dius  R'  and  apothem  r  of  the  second  polygon. 

Now  OH  is  the  arithmetic  mean  between  OC  and 
OD,  and  OA'  is  the  mean  proportional  between  OC 
and  OD'. 


137.  Now,  take  on  <9C,  OE=OA'  and  draw  ^'^. 
Then  A'  H  being  less  than  ^'C,  and  /  D'A'C  being 
bisected  by  A'E, 

ED'  is  less  than  \CH  ,  i.  e.,  less  than  \CD 

.•.   R\  —  r\  is  less  than  \{R  —  *")• 
As  the  number  of  sides  is  increased,  the  polygon 
approaches  the  circle  of  the  same  perimeter,  and  R 
and  r  approach  the  radius  of  the  circle. 
That  is, 


=  the  diameter  of  the  circle  =  — . 

7t 

Also, 

_/?]==  Rr\  or  R'   _-  =-f?\ 


and  ~  =  -^,  and  so  on. 


Multiplying  both  sides, 
'     -  '     -  '  *    •  •  -  -  —  tne  radius  of  the  circle  =  —-. 


IN  PAPER  FOLDING  75 

138.  The  radius  of  the  circle  lies  between  Rn  and 
rnJ  the  sides  of  the  polygon  being  4-2"  in  number; 

2  2 

and  TT  lies  between—  and  •-  -.      The  numerical  value 
rn  Rn 

of  n  can  therefore  be  calculated  to  any  required  de 
gree  of  accuracy  by  taking  a  sufficiently  large  number 
of  sides. 

The  following  are  the  values  of  the  radii  and  ap- 
othems  of  the  regular  polygons  of  4,  8,  16....  2048 
sides. 

4-gon,   r  =  0-500000     R  =  r\/^=  0  -707107 
8-gon,   n  =  0  -603553    RI  =  0-653281 

2048-gon,   r<>  =  0-636620    ^9  =  0-636620. 


139.   If  R"  be  the  radius  of  a  regular  isoperimetric 
polygon  of  4^  sides 

_2_ 

or  in  general 


T~\  ^^ 


140.  The  radii  R\,   R^. ....  successively  diminish,  ^ 

and  the  ratio  -77-12  less  than   unity  and   equal  to   the 
cosine  of  a  certain  angle  a. 


RZ           \l  +  cos  a  a 

^  =  \ 2--=cos2- 


76  GEOMETRIC  EXERCISES 

R  i,\  ,  of. 


multiplying  together  the  different  ratios,  we  get 
J^jt+l  =J?i'Cosa-  cos  —  •  cos 


The  limit  of  cos  or-cos  —  ?  .  .  .  .  cos         ,  when  £  =  oo, 

O  ^  « 

is  —  -  -  ,  a  result  known  as  Euler's  Formula. 

141.  It  was  demonstrated  by  Karl  Friedrich  Gauss* 
(1777-1855)  that  besides  the  regular  polygons  of  2", 
3-2",  5-2",  15  -2"  sides,  the  only  regular  polygons 
which  can  be  constructed  by  elementary  geometry 
are  those  the  number  of  whose  sides  is  represented 
by  the  product  of  2"  and  one  or  more  different  num 
bers  of  the  form  2"*-f-l.  We  shall  show  here  how 
polygons  of  5  and  17  sides  can  be  described. 
The  following  theorems  are  required  :f 
(1)  If  (7  and  Z>  are  two  points  on  a  semi-circum 
ference  ACDB,  and  if  C'  be  symmetric  to  C  with  re 
spect  to  the  diameter  AB,  and  R  the  radius  of  the 
circle, 


. 
AC-BC=R-CC  .............  iii. 

(2)   Let  the  circumference  of  a  circle  be  divided 
into  an  odd  number  of  equal  parts,  and  let  AO  be  the 

*Beman  and  Smith's  translation  of  Fink's  History  of  Mathematics,  p. 
245;  see  also  their  translation  of  Klein's  Famous  Problems  of  Elementary 
Geometry  •,  pp.  16,  24,  and  their  New  Plane  and  Solid  Geometry,  p.,  212. 

t  These  theorems  may  be  found  demonstrated  in  Catalan's  Theortmes  et 
Problemes  de  Geomttrie  Elementaire. 


IN  PAPER  FOLDING 


77 


diameter  through  one  of  the  points  of  section  A  and 
the  mid-point  O  of  the  opposite  arc.  Let  the  points 
of  section  on  each  side  of  the  diameter  be  named  A\, 
At,  As.  .  .  .An,  and  A'i,  A'2,  A'3.  .  .  .A'n  beginning  next 
to  A. 

Then  OAi  •  OA*  •  OA3  ____  OAn  =  Rn  ......  iv. 

n 

and  OAi  -  OA^  •  OA±  ....  OA  =  R*. 


142.  It  is  evident  that  if  the  chord  OAn  is  deter 
mined,  the  angle  AnOA  is  found  and  it  has  only  to  be 
divided  into  2"  equal  parts,  to  obtain  the  other  chords. 


i<<.   Let  us  first  take  the  pentagon. 
By  theorem  iv, 


By  theorem  i, 

R(OAl—  OA2)= 


GEOMETRIC  EXERCISES 


1), 


and    t?^2  =      (i/5  —  1). 

Hence  the  following  construction. 

Take   the   diameter  ^C6>,    and   draw  the  tangent 
AF.     Take  D  the   mid-point   of   the   radius    OC  and 


On  OC  as  diameter  describe  the  circle  AE'CE. 
Join  FD  cutting  the  inner  circle  in  E  and  £'. 
Then  FE'=OA,  and  FE=  OA2. 

144.   Let  us  now  consider  the  polygon  of  seven 
teen  sides. 
Here* 

Az  •  OA±  •  OA6  -  OA«  -OAr  OA%  =  R*. 


and    O 

By  theorems  i.  and  ii. 
OAl  -  OA±  = 


Suppose 


*The  principal  steps  are  given.  For  a  full  exposition  see  Catalan's  Thto- 
rtmes  et  Probltmes  de  Gtomttrie  Elententaire.  The  treatment  is  given  in  full 
in  Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of  Elementary 
Geometry,  chap.  iv. 


IN  PAPER  FOLDING  79 


Then  MN=R*  and  PQ  =  R*. 
Again  by  substituting  the  values  of  M,  IV,  P  and 
Q  in  the  formulas 

MN=R\    PQ=R^ 
and  applying  theorems  i.  and  ii.  we  get 
(M—  N)~  (P—  Q^)  =  R. 

ALSO  by  substituting  the  values  of  M,  N,  P  and  Q  in 
the  above  formula  and   applying  theorems  i.  and  ii. 

we  get 

(  M—  N}  (P—  Q}  =  4^2. 

Hence  M—N,  P—  Q,  J/,  N,  P  and  Q  are  deter 
mined. 
Again 


Hence  OAS  is  determined. 

145.  By  solving  the  equations  we  get 

M—  N=  ^R  (1  -f  1/17). 
P  —  Q  =  ±R(—  1  +  1/17). 


OA8  =  IR\_— 1+1/17  +  1/34  —  2i/17 

—  2  iXl7+3i/17+  1/170— 26  V 11— 4  |/34+2v/17  ] 
=  J7?[—  1  +  /I7  + 1/34  —  21/17 

+  31/17  — 1/170  +  38J/17], 


8o 


GEOMETRIC  EXERCISES 


146.  The  geometric  construction  is  as  follows : 

Let  BA  be  the  diameter  of  the  given  circle ;    O  its 

center.    Bisect  OA  in  C.    Draw  AD  at  right  angles  to 

OA  and  take  AD  =  AB.     Draw  CD.     Take  E  and  E' 

in  CD  and  on  each  side  of  C  so  that  C£=  CE'  =  CA. 


Fig.  51. 

Bisect  ^Z>  in  G  and  .£'£>  in  G'.  Draw  Z>^  per 
pendicular  to  CD  and  take  DF=  OA. 

Draw  FG  and  /r£'. 

Take  H  in  7^7  and  H'  in  ^£'  produced  so  that 
GH=EG  and  £'.#'  ==  G'D. 

Then  it  is  evident  that 


IN  PAPER  FOLDING.  81 

also 


FH'  =  P,    '.-  (FH'  —  DE')  FH=  DF*  =  &. 

Again  in  DF  take  A'  such  that  FK=FH. 

Draw  KL  perpendicular  to  DF  and  take  L  in  KL 
such  that  FL  is  perpendicular  to  DL. 

Then  FL*  =DF-  FK=RN. 

Again  draw  J7W  perpendicular  to  FH'  and  take 
H'N=  FL.  Draw  NM  perpendicular  to  NH'  .  Find 
M  in  NM  such  that  ZT'J/  is  perpendicular  to  FM. 
Draw  MF'  perpendicular  to  FH'  . 

Then 

F'H'  '  FF'  =  ^'  J/2  « 


But  FF' 


XII.   GENERAL  PRINCIPLES. 

147.  In  the  preceding  pages  we  have  adopted  sev 
eral  processes,   e.  g.,    bisecting   and   trisecting   finite 
lines,  bisecting  rectilineal  angles   and  dividing  them 
into   other  equal  parts,  drawing  perpendiculars  to  a 
given  line,  etc.      Let  us  now  examine   the  theory  of 
these  processes. 

148.  The  general  principle  is  that  of  congruence. 
Figures  and  straight  lines  are  said  to  be  congruent,  if 
they  are  identically  equal,  or  equal  in  all  respects. 

In  doubling  a  piece  of  paper  upon  itself,  we  ob 
tain  the  straight  edges  of  two  planes  coinciding  with 
each  other.  This  line  may  also  be  regarded  as  the 
intersection  of  two  planes  if  we  consider  their  posi 
tion  during  the  process  of  folding. 

In  dividing  a  finite  straight  line,  or  an  angle  into  a 
number  of  equal  parts,  we  obtain  a  number  of  con 
gruent  parts.  Equal  lines  or  equal  angles  are  con 
gruent. 

149.  Let  X'X  be   a  given  finite  line,  divided  into 
any  two  parts  by  A'.     Take  O  the  mid-point  by  doub 
ling  the  line  on  itself.    Then  OA'  is  half  the  difference 


PAPER  FOLDING  83 

between  A'X  and  X'A'.  Fold  X'X  over  O,  and  take 
A  in  OX  corresponding  to  A'.  Then  A  A'  is  the  differ 
ence  between  A'X  and  Jf'^'  and  it  is  bisected  in  O. 

I 1 1 1 1 

X'  A'  O  A  X 

Fig.  52- 

As  -4'  is  taken  nearer  O,  A' O  diminishes,  and  at  the 
same  time  A' A  diminishes  at  twice  the  rate.  This 
property  is  made  use  of  in  finding  the  mid-point  of  a 
line  by  means  of  the  compasses. 

150.  The    above    observations    apply   also    to    an 
angle.      The  line  of  bisection  is  found  easily  by  the 
compasses  by  taking  the  point  of  intersection  cf  two 
circles. 

151.  In   the   line   X'X,    segments   to   the   right   of 
0  may  be  considered  positive  and  segments,  to  the 
eft  of   O  may  be   considered   negative.      That   is,    a 

point  moving  from  O  to  A  moves  positively,  and  a 
point  moving  in  the  opposite  direction  OA'  moves 

negatively. 

AX=OX—OA. 

O  A' =  OX'  — A'X', 
both  members  of  the  equation  being  negative.* 

152.  If  OA,  one  arm  of  an  angle  A  OP,  be  fixed  and 
OP  be   considered   to   revolve   round    O,    the   angles 
which  it  makes  with  OA  are  of  different  magnitudes. 

*See  Beman  and  Smith's  New  Plane  and  Solid  Geometry,  p.  56. 


84  GEOMETRIC  EXERCISES 

All  such  angles  formed  by  OP  revolving  in  the  direc 
tion  opposite  to  that  of  the  hands  of  a  watch  are  re 
garded  positive.  The  angles  formed  by  OP  revolving 
in  an  opposite  direction  are  regarded  negative.* 

153.  After  one  revolution,  OP  coincides  with  OA. 
Then   the   angle   described  is  called  a  perigon,  which 
evidently  equals   four   right   angles.      When    OP  has 
completed   half   the   revolution,    it   is   in   a   line   with 
OAB.      Then  the  angle  described  is  called  a  straight 
angle,    which     evidently    equals     two    right    angles. f 
When  OP  has  completed  quarter  of  a  revolution,  it  is 
perpendicular  to  OA.      All  right   angles  are  equal  in 
magnitude.      So   are   all   straight   angles   and  all  peri- 
gons. 

154.  Two   lines  at   right  angles  to  each  other  form 
four  congruent   quadrants.      Two  lines  otherwise   in 
clined  form  four  angles,  of  which   those  vertically  op 
posite  are  congruent. 

155.  The  position   of  a  point  in  a  plane   is  deter 
mined  by  its  distance  from  each  of  two  lines  taken  as 
above.      The  distance  from  one  line  is  measured  par 
allel  to  the  other.      In   analytic  geometry  the  proper 
ties  of  plane  figures  are  investigated  by  this  method. 
The  two  lines  are  called   axes ;   the  distances  of  the 
point  from   the  axes  are  called  co-ordinates,  and  the 
intersection   of   the  axes  is  called   the   origin.      This 

*  See  Beman  and  Smith's  New  Plane  and  Solid  Geometry,  p.  56. 
t/*.,p.  5- 


IN  PAPER  FOLDING.  85 

method  was  invented  by  Descartes  in  1637  A.  D.*     It 
has  greatly  helped  modern  research. 

156.  If  X'X,   YY'  be   two   axes  intersecting  at  O, 
distances  measured  in  the  direction   of  OX,  i.  e.,  to 
the  right   of  O  are  positive,  while  distances  measured 
to  the  left  of  O  are  negative.      Similarly  with  reference 
to  YY',  distances  measured  in  the  direction  of  6>Fare 
positive,  while   distances  measured  in  the  direction  of 
OY'  are  negative. 

157.  Axial  symmetry  is  defined  thus  :   If  two  fig 
ures  in  the  same  plane  can  be  made  to  coincide  by 
turning  the  one  about  a  fixed  line  in  the  plane  through 
a  straight  angle,  the  two  figures  are   said  to  be  sym 
metric  with  regard  to  that  line  as  axis  of  symmetry. f 

158.  Central  symmetry  is  thus  defined :   If  two  fig 
ures  in  the  same  plane  can  be  made  to  coincide  by 
turning    the    one    about  a   fixed   point    in  that  plane 
through  a  straight  angle,  the  two  figures  are  said  to 
be  symmetric  with  regard  to  that  point  as  center  of 
symmetry.  J 

In  the  first  case  the  revolution  is  outside  the  given 
plane,  while  in  the  second  it  is  in  the  same  plane. 

If  in  the  above  two  cases,  the  two  figures  are  halves 
of  one  figure,  the  whole  figure  is  said  to  be  symmetric 
with  regard  to  the  axis  or  center — these  are  called  axis 
or  center  of  symmetry  or  simply  axis  or  center. 

*Beman  and  Smith's  translation  of  Fink's  History  of  Mathematics,  p.  230. 
t  Beman  and  Smith's  New  Plane  and  Solid  Geometry,  p.  26. 
t/£.,p.  183. 


86  GEOMETRIC  EXERCISES 

159.  Now,  in  the  quadrant  XOVmake  a  triangle 
PQR.  Obtain  its  image  in  the  quadrant  VOX'  by 
folding  on  the  axis  YY'  and  pricking  through  the 
paper  at  the  vertices.  Again  obtain  images  of  the  two 
triangles  in  the  fourth  and  third  quadrants.  It  is  seen 
that  the  triangles  in  adjacent  quadrants  posses  axial 


Fig-  53- 

symmetry,  while  the  triangles  in  alternate  quadrants 
possess  central  symmetry. 

160.  Regular  polygons  of  an  odd  number  of  sides 
possess  axial  symmetry,  and  regular  polygons  of  an 
even  number  of  sides  possess  central  symmetry  as 
well. 


AV  PAPER  FOLDING.  87 

161.  If  a  figure  has  two  axes  of  symmetry  at  right 
angles  to  each  other,  the  point  of  intersection  of  the 
axes  is  a  center  of  symmetry.  This  obtains  in  reg 
ular  polygons  of  an  even  number  of  sides  and  certain 
curves,  such  as  the  circle,  ellipse,  hyperbola,  and  the 
lemniscate ;  regular  polygons  of  an  odd  number  of 


Fig.  54- 

sides  may  have  more  axes  than  one,  but  no  two  of 
them  will  be  at  right  angles  to  each  other.  If  a  sheet 
of  paper  is  folded  double  and  cut,  we  obtain  a  piece 
which  has  axial  symmetry,  and  if  it  is  cut  fourfold,  we 
obtain  a  piece  which  has  central  symmetry  as  well,  as 
in  Fig.  54. 


88  GEOMETRIC  EXERCISES 

162.  Parallelograms  have  a  center  of  symmetry. 
A  quadrilateral  of  the  form  of  a  kite,  or  a  trapezium 
with  two  opposite  sides  equal  and  equally  inclined  to 
either  of   the   remaining  sides,    has  an   axis  of   sym 
metry. 

163.  The  position   of  a  point  in  a  plane  is  also  de 
termined  by  its  distance  from  a  fixed  point  and  the 
inclination  of  the  line  joining  the  two  points  to  a  fixed 
line  drawn  through  the  fixed  point. 

If  OA  be  the  fixed  line  and  P  the  given  point,  the 
length  OP  and  /_AOP,  determine  the   position  of  P. 


FiR.  55- 

O  is  called  the  pole,  OA  the  prime-vector,  OP  the 
radius  vector,  and  /_AOP  the  vectorial  angle.  OP 
and  ^_AOP  are  called  polar  co-ordinates  of  P. 

164.  The  image  of  a  figure  symmetric  to  the  axis 
OA  may  be  obtained  by  folding  through  the  axis  OA. 
The  radii  vectores  of  corresponding  points  are  equally 
inclined  to  the  axis. 

165.  Let  ABC  be  a  triangle.     Produce  the  sides 
CA,   AB,   BC  to  Z>,   E,  F  respectively.     Suppose  a 
person  to  stand  at  A  with  face  towards  D  and  then  to 


IN  PAPER  FOLDING.  8g 

proceed  from  A  to  B,  B  to  C,  and  C  to  A.  Then  he 
successively  describes  the  angles  DAB,  EBC,  FCD. 
Having  come  to  his  original  position  A,  he  has  corn- 


Fig.  56. 

pleted  a  perigon,  i.  e. ,  four  right  angles.  We  may 
therefore  infer  that  the  three  exterior  angles  are  to 
gether  equal  to  four  right  angles. 

The  same  inference  applies  to  any  convex  polygon. 

161.  Suppose  the  man  to  stand  at  A  with  his  face 
towards  C,  then  to  turn  in  the  direction  of  AB  and 
proceed  along  AB,  BC,  and  CA. 

In  this  case,  the  man  completes  a  straight  angle, 
i.  e.,  two  right  angles.  He  successively  turns  through 
the  angles  CAB,  EBC,  and  FCA.  Therefore  /_EBF 
+  Z  FCA  -f  /  CAB  (neg.  angle)  =  a  straight  angle. 

This  property  is  made  use  of  in  turning  engines 
on  the  railway.  An  engine  standing  upon  DA  with 
its  head  towards  A  is  driven  on  to  CF,  with  its  head 
towards  F.  The  motion  is  then  reversed  and  it  goes 
backwards  to  EB.  Then  it  moves  forward  along  BA 
on  to  AD.  The  engine  has  successively  described 


90  GEOMETRIC  EXERCISES 

the  angles  ACB,  CBA,  and  BAG.  Therefore  the 
three  interior  angles  of  a  triangle  are  together  equal 
to  two  right  angles. 

167.  The  property  that  the  three  interior  angles  of 
a  triangle  are  together  equal  to  two  right  angles  is 
illustrated  as  follows  by  paper  folding. 

Fold  CC'  perpendicular  to  AB.  Bisect  C  B  in  Nt 
and  AC'  in  M.  Fold  NA ',  MB'  perpendicular  to  AB, 
meeting  BC  and  AC'm  A'  and  B'.  Draw  A' C ,  B' C . 


By  folding  the  corners  on  NA' t  MB'  and  A'B',  we 
find  that  the  angles  A,  B,  C  of  the  triangle  are  equal 
to  the  angles  B'C'A,  BC'A',  and  A' C  B'  respectively, 
which  together  make  up  two  right  angles. 

168.  Take  any  line  ABC,  Draw  perpendiculars 
to  ABC  at  the  points  A,  B,  and  C.  Take  points 
Z>,  E,  F'm  the  respective  perpendiculars  equidistant 


IN  PAPER  FOLDING.  91 

from  their  feet.  Then  it  is  easily  seen  by  superposi 
tion  and  proved  by  equal  triangles  that  DE  is  equal 
to  AB  and  perpendicular  to  AD  and  BE,  and  that 
EF'is  equal  to  BC  and  perpendicular  to  BE  and  CF. 
Now  AB  (=DE}  is  the  shortest  distance  between  the 
lines  AD  and  BE,  and  it  is  constant.  Therefore  AD 


Fig.  58. 

and  BE  can  never  meet,  i.  e.,  they  are  parallel.  Hence 
lines  which  are  perpendicular  to  the  same  line  are 
parallel. 

The  two  angles  BAD  and  EBA  are  together  equal 
to  two  right  angles.  If  we  suppose  the  lines  AD  and 
BE  to  move  inwards  about  A  and  B,  they  will  meet 
and  the  interior  angles  will  be  less  than  two  right 
angles.  They  will  not  meet  if  produced  backwards. 
This  is  embodied  in  the  much  abused  twelfth  postulate 
of  Euclid's  Elements.* 

169.  If  AGffbe  any  line  cutting  BE  in  G  and  CF 
in  H,  then 


*For  historical   sketch   see  Beman   and   Smith's  translation  of  Fink's 
History  of  Mathematics,  p.  270. 


92  GEOMETRIC  EXERCISES 

/  GAD=  the  alternate  /_AGB, 
•.•  each  is  the  complement  of  /_BAG;  and 
/_HGE—  the  interior  and  opposite  /  GAD. 
.  • .   they  are  each  =  /  A  GB. 

Also  the  two  angles  GAD  and  EGA  are  together 
equal  to  two  right  angles. 

170.  Take  a  line  AX  and  mark  off  on  it,  from  A, 
equal  segments  AB,  BC,  CD,  DE..  ..Erect  perpen 
diculars  to  AE  at  B,  C,  D,  E. .  . .  Let  a  line  AF'  cut 
the  perpendiculars  in  B' ,  C ',  D' ,  E' . .  . .  Then  AB', 
B'C',  CD',  D'E' ..  .  .are  all  equal. 


A         B         C         D         E         F 

Fig.  59- 

If  AB,  BC,   CD,  DE  be  unequal,  then 
AB:BC=AB':B'C 
BC:  CD  =  B'C\  CD',  and  so  on. 

171.  If  ABCDE. .  . .  be  a  polygon,  similar  polygons 
may  be  obtained  as  follows. 

Take  any  point  O  within  the  polygon,  and  draw 
OA,  OB,  OC,.... 

Take  any  point  A'  in  OA  and  draw  A'B',  B'C', 
C'D, parallel  to  AB,  BC,  CD respectively. 


IN  PAPER  FOLDING. 


93 


Then    the    polygon   A' B' C D' will    be    similar  to 

ABCD . .  . .  The  polygons  so  described  around  a  com 
mon  point  are  in  perspective.  The  point  O  may  also 
lie  outside  the  polygon.  It  is  called  the  center  of  per 
spective. 

172.  To  divide  a  given  line  into  2,  3,  4,  5. .  .  .equal 
parts.     Let  AB  be   the  given  line.     Draw  AC,  BD 


at  right  angles  to  AB  on    opposite    sides  and  make 
AC=BD.      Draw  CD  cutting  AB  in  P*.      Then 


Now  produce  AC  and  take  CE  =  EF=  FG  .  .  .  . 
^AC  or  BD.  Draw  DE,  DF,  DG  ____  cutting  AB 
in  PS,  -A,  A,  .... 

Then  from  similar  triangles, 


94  GEOMETRIC  EXERCISES 

.-.  P9.B:  AB  =  BD:  AF 

=  1  :3. 
Similarly 


and  so  on. 

If  AB  =  \, 


P  - 

A  =  3-.  4, 


*(«  +  *) 

But  A  Pi  -\-  P2P3  +  ^  A  -f is  ultimately  ==  AB. 


Or 


1        1  _      1 
2"  ""  "3  =  2~- 3 ' 


1        1  1 


n        n-\-\       n(n-\- 1) 
Adding 


F2  +  ^3  +  "-'  + 


IN  PAPER  FOLDING.  95 


±        J_ 

'      -"1""        *" 


The  limit  of  1  --  when  n  is  co  is  1. 


173.  The    following    simple    contrivance    may   be 
used  for  dividing  a  line  into  a  number  of  equal  parts. 

Take  a  rectangular  piece  of  paper,  and  mark  off  n 
equal  segments  on  each  or  one  of  two  adjacent  sides. 
Fold  through  the  points  of   section    so  as  to  obtain 
perpendiculars  to  the  sides.      Mark  the  points  of  sec 
tion  and  the  corners  0,  1,  2,  ......     Suppose  it  is  re 

quired  to  divide  the  edge  of  another  piece  of  paper 
AB  into  n  equal  parts.  Now  place  AB  so  that  A  or 
B  may  lie  on  0,  and  B  or  A  on  the  perpendicular 
through  n. 

In  this  case  AB  must  be  greater  than  ON.  But 
the  smaller  side  of  the  rectangle  may  be  used  for 
smaller  lines. 

The  points  where  AB  crosses  the  perpendiculars 
are  the  required  points  of  section. 

174.  Center  of  mean  position.      If  a  line  AB  con 
tains  (m  4-  »)  equal  parts,   and  it  is  divided  at   C  so 
that  AC  contains  m  of  these  parts  and  CB  contains  n 
of  them  ;   then  if  from  the  points  A,  C,  B  perpendicu 
lars  AD,  CF,  BE  be  let  fall  on  any  line, 

m-BE  -f  n-AZ>  =  (m  -f  «)  •  CF. 

Now,  draw  BGH  parallel  to  ED  cutting  CFin  G 
and  AD  in  H.  Suppose  through  the  points  of  divi 
sion  AB  lines  are  drawn  parallel  to  BH.  These  lines 


g5  GEOMETRIC   EXERCISES 

will  divide  AH  into  (m-{-ri)  equal  parts  and  CG  into 
n  equal  parts. 


and  since  DH  and  BE  are  each  =  GF, 


Hence,  by  addition 

n-l  -f  Z>  +  m-BE  =  (m  +  «) 
»  •  ^4Z>  +  0*  •  .#£•  —  (/»  -f  »)  •  CF. 

C  is  called  the  center  of  mean  position,  or  the 
mean  center  of  A  and  B  for  t.he  system  of  multiples 
m  and  n. 

The  principle  can  be  extended  to  any  number  of 
points,  not  in  a  line.  Then  if  P  represent  the  feet  of 
the  perpendiculars  on  any  line  from  A,  B,  C,  etc.,  if 
a,  b,  c  ...be  the  corresponding  multiples,  and  if  M 
be  the  mean  center 

c-CP.... 


If  the  multiples  are  all  equal  to  a,  we  get 

a(AP+BP+CP+..  ..}=na-MP 
n  being  the  number  of  points. 

175.  The  center  of  mean  position  of  a  number  of 
points  with  equal  multiples  is  obtained  thus.  Bisect 
the  line  joining  any  two  points  A,  B  in  G,  join  G  to  a 
third  point  C  and  divide  GC  in  71  so  that  GH=\GC\ 
join  If  to  a  fourth  point  D  and  divide  HD  in  K  so  that 
HK=\HD  and  so  on:  the  last  point  found  will  be 
the  center  of  mean  position  of  the  system  of  points. 


IN  PAPER  FOLDING.  97 

176.  The  notion  of  mean  center  or  center  of  mean 
position  is  derived  from  Statics,  because  a  system  of 
material  points  having  their  weights  denoted  by  a,  b, 
c  .  .  .  .  ,  and  placed  at  A,  B,  C  .  .  .  .  would  balance  about 
the  mean  center  M,  if  free  to  rotate  about  M  under 
the  action  of  gravity. 

The  mean  center  has  therefore  a  close  relation  to 
the  center  of  gravity  of  Statics. 

177.  The  mean  center  of  three  points  not  in  a  line, 
is  the  point  of  intersection  of  the  medians  of  the  tri 
angle  formed  by  joining  the  three  points.   This  is  also 
the   center   of   gravity  or   mass  center  of  a  thin  tri 
angular  plate  of  uniform   density. 

178.  If  M  is  the  mean  center  of  the  points  A,  B, 
C,  etc.,  for  the  corresponding  multiples  a,  b,  c,  etc., 
and  if  P  is  any  other  point,  then 


=  a  •  A  M*  +  b  •  B  M*  +  c-  CM  2  +  .  .  .  . 


Hence  in  any  regular  polygon,  if  O  is  the  in-center 
or  circum-center  and  P  is  any  point 

4-  BP*  +  ....=  OA*  +  OB^  +  ....+  »•  OP2 


Now 

AB^ 
Similarly 


98  GE  OME  TRIG  EXER  CISES 

Adding 


179.  The  sum  of  the  squares  of  the  lines  joining 
the  mean  center  with  the  points  of  the  system  is  a 
minimum. 

If  J/be  the  mean  center  and  P  any  other  point 
not  belonging  to  the  system, 

2P^  =  2MA2+2PM'*,  (where  2  stands  for  "the 
sum  of  all  expressions  of  the  type"). 

.•.  2PA2  is  the  minimum  when  PAf=Q,  i.  e., 
when  P  is  the  mean  center. 

180.  Properties   relating  to  concurrence   of  lines 
and  collinearity  of  points  can  be  tested  by  paper  fold 
ing.*     Some  instances  are  given  below: 

(1)  The  medians  of  a  triangle  are  concurrent.   The 
common  point  is  called  the  centroid. 

(2)  The   altitudes   of   a   triangle    are    concurrent 
The  common  point  is  called  the  orthocenter. 

(3)  The  perpendicular  bisectors  of  the  sides  of  a 
triangle  are  concurrent.     The  common  point  is  called 
the  circum-center. 

(4)  The  bisectors  of  the  angles  of  a  triangle  are 
concurrent.    The  common  point  is  called  the  in-center. 

(5)  Let  ABCD  be    a   parallelogram   and  P  any 
point.     Through  P  draw  G£T  and  EF  parallel  to  BC 

*For  treatment  of  certain  of  these  properties  see  Beman  and  Smith's 
Neiu  Plane  and  Solid  Geometry,  pp.  84,  182. 


IN  PAPER  FOLDING.  99 

and  AB  respectively.     Then  the  diagonals  EG,  HF, 
and  the  line  DB  are  concurrent. 

(6)  If  two  similar  unequal  rectineal  figures  are  so 
placed  that  their  corresponding  sides  are  parallel,  then 
the  joins   of   corresponding   corners    are    concurrent. 
The  common  point  is  called  the  center  of  similarity. 

(7)  If  two  triangles  are  so  placed  that  their  corners 
are  two  and  two  on  concurrent  lines,  then  their  corre 
sponding  sides  intersect  collinearly.     This  is  known 
as  Desargues's  theorem.      The  two  triangles  are  said 
to  be  in  perspective.     The  point  of  concurrence  and 
line  of   collinearity  are  respectively  called  the  center 
and  axis  of  perspective. 

(8)  The  middle  points  of  the  diagonals  of  a  com 
plete  quadrilateral  are  collinear. 

(9)  If  from  any  point  on  the  circumference  of  the 
circum-circle  of  a  triangle,  perpendiculars  are  dropped 
on   its   sides,    produced   when   necessary,    the  feet  of 
these  perpendiculars  are  collinear.    This  line  is  called 
Simson's  line. 

Simson's  line  bisects  the  join  of  the  orthocenter 
and  the  point  from  which  the  perpendiculars  are 
drawn. 

(10)  In  any  triangle  the  orthocenter,  circum-center, 
and  centroid  are  collinear. 

The  mid-point  of  the  join  of  the  orthocenter  and 
circum-center  is  the  center  of  the  nine-points  circle,  so 
called  because  it  passes  through  the  feet  of  the  alti 
tudes  and  medians  of  the  triangle  and  the  mid-point 


ioo  GEOMETRIC  EXERCISES 

of  that  part  of  each  altitude  which  lies  between  the 
orthocenter  and  vertex. 

The  center  of  the  nine-points  circle  is  twice  as  far 
from  the  orthocenter  as  from  the  centroid.  This  is 
known  as  Poncelet's  theorem. 

(11)  If  A,  B,  C,  D,  E,  F,  are  any  six  points  on  a 
circle  which  are  joined  successively  in  any  order,  then 
the  intersections  of  the  first  and  fourth,  of  the  second 
and  fifth,  and  of  the  third  and  sixth  of  these  joins  pro 
duced  when  necessary)  are  collinear.      This  is  known 
as  Pascal's  theorem. 

(12)  The  joins  of  the  vertices  of  a  triangle  with  the 
points  of  contact  of  the  in-circle  are  concurrent.     The 
same  property  holds  for  the  ex- circles. 

(13)  The  internal  bisectors  of  two  angles  of  a  tri 
angle,  and  the  external  bisector  of  the  third  angle  in 
tersect  the  opposite  sides  collinearly. 

(14)  The  external  bisectors  of  the  angles  of  a  tri 
angle  intersect  the  opposite  sides  collinearly. 

(15)  If  any  point  be  joined  to  the  vertices   of   a 
triangle,   the  lines  drawn  through  the   point   perpen 
dicular  to   those  joins  intersect  the  opposite  sides  of 
the  triangle  collinearly. 

(16)  If  on   an   axis   of  symmetry  of  the  congruent 
triangles  ABC,  A'fi'C  a  point  O  be  taken  A'O,  B' O, 
and  CO  intersect  the  sides  BC,  CA  and  AB  collin 
early. 

(17)  The  points  of  intersection  of  pairs  of  tangents 
to  a  circle  at  the  extremities  of  chords  which  pass 


IN  PAPER  FOLDING  101 

through  a  given  point  are  collinear.   This  line  is  called 
the  polar  of  the  given  point  with  respect  to  the  circle. 

(18)  The  isogonal  conjugates  of  three  concurrent 
lines  AX,  BX,  CX  with  respect  to  the  three  angles  of 
a  triangle  ABC  are  concurrent.      (Two  lines  AX,  AY 
are  said  to  be  isogonal  conjugates  with   respect  to  an 
angle  BAC,   when   they  make   equal  angles  with  its 
bisector.) 

(19)  If  in  a  triangle  ABC,  the  lines  AA\  BB',  CC 
drawn  from   each  of  the  angles  to  the  opposite  sides 
are  concurrent,  their  isotomic  conjugates  with  respect 
to  the  corresponding  sides  are  also  concurrent.     (The 
lines  A  A',  A  A"  are  said  to  be  isotomic  conjugates, 
with  respect  to  the  side  BC  of  the  triangle  ABC,  when 
the  intercepts  BA'  and  CA"  are  equal.) 

(20)  The  three  symmedians  of  a  triangle  are  con 
current.     (The  isogonal  conjugate  of  a  median  AM oi 
a  triangle  is  called  a  symmedian.) 


XIII.   THE  CONIC  SECTIONS. 

SECTION  I.— THE  CIRCLE. 

181.  A  piece  of  paper  can   be  folded  in  numerous 
ways  through  a  common  point.    Points  on  each  of  the 
lines  so  taken  as  to  be  equidistant  from  the  common 
point  will  lie  on  the  circumference  of  a  circle,  of  which 
the  common  point  is  the  center.      The  circle  is  the 
locus   of   points   equidistant   from   a  fixed   point,   the 
centre. 

182.  Any    number    of    concentric    circles    can    be 
drawn.     They  cannot  meet  each  other. 

183.  The  center  may  be  considered  as  the  limit  of 
concentric  circles  described   round   it  as  center,   the 
radius  being  indefinitely  diminished. 

184.  Circles  with   equal   radii   are   congruent   and 
equal. 

185.  The  curvature  of  a  circle  is  uniform  through 
out  the  circumference.   A  circle  can  therefore  be  made 
to  slide  along  itself  by  being  turned  about  its  center. 
Any  figure  connected  with  the  circle  may  be  turned 
about  the  center  of  the  circle  without  changing  its  re 
lation  to  the  circle. 


PAPER  FOLDING  103 

186.  A  straight  line  can  cross  a  circle  in  only  two 
points. 

187.  Every  diameter  is  bisected  at   the  center  of 
the  circle.      It  is  equal  in  length  to  two  radii.      All 
rjiameters,  like  the  radii,  are  equal. 

188.  The  center  of  a  circle  is  its  center  of  sym 
metry,  the  extremities  of  any  diameter  being  corre 
sponding  points. 

189.  Every  diameter  is  an  axis  of  symmetry  of  the 
circle,  and  conversely. 

190.  The  propositions  of  §§  188,  189  are  true  for 
systems  of  concentric  circles. 

191.  Every   diameter   divides   the   circle    into   two 
equal  halves  called  semicircles. 

192.  Two  diameters  at   right  angles  to  each  other 
divide  the  circle  into  four  equal  parts  called  quadrants. 

193.  By  bisecting  the   right   angles   contained   by 
the  diameters,  then   the  half  right  angles,  and  so  on, 

we  obtain  2n  equal  sectors  of  the  circle.      The  angle 

4. 
between  the  radii  of  each  sector  is  — -  of  a  right  angle 

27T  7t 

Of    . 

2*  —  2»-i- 

194.  As  shown  in  the  preceding  chapters,  the  right 
angle  can  be  divided  also  into  3,  5,  9,  10,  12,  15  and 
17  equal  parts.     And  each  of  the  parts  thus  obtained 
can  be  subdivided  into  2"  equal  parts. 


io4  GEOMETRIC  EXERCISES 

195.  A  circle  can  be  inscribed  in  a  regular  polygon, 
and  a  circle  can  also  be  circumscribed  round  it.     The 
former  circle  will  touch  the  sides  at  their  mid-points. 

196.  Equal  arcs  subtend  equal  angles  at  the  cen 
ter;  and  conversely.     This  can  be  proved  by  super 
position.      If  a  circle  be  folded  upon  a  diameter,  the 
two  semicircles  coincide.      Every  point  in  one  semi- 
circumference  has  a  corresponding  point  in  the  other, 
below  it. 

197.  Any  two  radii  are  the  sides  of  an  isosceles  tri 
angle,  and  the  chord  which  joins  their  extremities  is 
the  base  of  the  triangle. 

198.  A  radius  which  bisects  the  angle  between  two 
radii  is  perpendicular  to  the  base  chord  and  also  bi 
sects  it. 

199.  Given    one   fixed    diameter,    any  number    of 
pairs  of  radii  may  be  drawn,  the  two  radii  of  each  set 
being  equally  inclined  to  the  diameter  on  each  side  of 
it.      The  chords  joining  the  extremities  of  each  pair  of 
radii  are  at  right  angles  to  the  diameter.     The  chords 
are  all  parallel  to  one  another. 

200.  The  same  diameter  bisects  all  the  chords  as 
well  as  arcs  standing  upon  the  chords,  i.  e.,  the  locus 
of  the  mid-points  of  a  system  of  parallel  chords  is  a 
diameter. 

201.  The  perpendicular  bisectors  of  all  chords  of  a 
circle  pass  through  the  center. 


IN  PAPER  FOLDING  105 

202.  Equal  chords  are  equidistant  from  the  center. 

203.  The  extremities  of  two  radii  which  are  equally 
inclined  to  a  diameter  on  each  side  of  it,  are  equi 
distant  from  every  point  in  the  diameter.      Hence,  any 
number  of  circles  can  be  described  passing  through 
the  two  points.     In  other  words,  the  locus  of  the  cen 
ters  of  circles  passing  through  two  given  points  is  the 
straight  line  which  bisects  at  right  angles  the  join  of 
the  points. 

204.  Let  CC'  be  a  chord  perpendicular  to  the  ra 
dius  OA.   Then  the  angles  AOCand  AOC'  are  equal. 
Suppose  both  move  on  the  circumference  towards  A 
with  the  same  velocity,  then  the  chord  CC'  is  always 
parallel  to  itself  and  perpendicular  to  OA.   Ultimately 
the  points  C,  A   and  C'  coincide   at  A,  and    CA'C'  is 
perpendicular  to  OA.     A  is  the  last  point  common  to 
the   chord   and   the   circumference.      CAC'  produced 
becomes  ultimately  a  tangent  to  the  circle. 

205.  The  tangent  is  perpendicular  to  the  diameter 
through  the  point  of  contact;  and  conversely. 

206.  If  two  chords  of  a  circle  are  parallel,  the  arcs 
joining  their  extremities  towards  the  same  parts  are 
equal.    So  are  the  arcs  joining  the  extremities  of  either 
chord  with  the  diagonally  opposite  extremities  of  the 
other  and  passing  through  the  remaining  extremities. 
This  is  easily  seen  by  folding  on  the  diameter  perpen 
dicular  to  the  parallel  chords. 


io6 


GEOMETRIC  EXERCISES 


207.  The  two  chords  and  the  joins  of  their  extrem 
ities  towards   the  same  parts  form  a  trapezoid  which 
has  an   axis  of  symmetry,  viz.,  the  diameter  perpen 
dicular  to  the  parallel  chords.      The  diagonals  of  the 
trapezoid  intersect  on  the  diameter.      It  is  evident  by 
folding  that  the  angles  between  each  of  the  parallel 
chords  and  each  diagonal  of  the  trapezoid  are  equal. 
Also  the  angles  upon  the  other  equal  arcs  are  equal. 

208.  The  angle  subtended  at  the  center  of  a  circle 
by  any  arc  is  double  the  angle  subtended  by  it  at  the 
circumference. 


Fig.  61. 


Fig.  62. 


Fig-  63. 


An   inscribed   angle   equals  half   the  central  angle 
standing  on  the  same  arc. 

Given 
A  VB  an  inscribed  angle,  and  AOB  the  central   angle 

on  the  same  arc.  AB. 
To  prove         that  /  A  VB  =  *-  /  A  OB. 
Proof. 

1.  Suppose    VO  drawn    through   center   O,   and   pro 
duced  to  meet  the  circumference  at  X. 


IN  PAPER  FOLDING  107 


Then 
2.  And          £XOB=  /.XVB  +  Z  VBO, 


3.  .-.  /_XVB  =  \_ 

4.  Similarly  /  A  VX=  \i_AOX  (each=zero  in  Fig.  62), 
and     .-.  LAVB  =  ±tAOB. 

The  proof  holds  for  all  three  figures,  point  A  hav 
ing  moved  to  X  (Fig.  62),  and  then  through  X  (Fig. 
63).* 

209.  The  angle   at  the  center  being  constant,  the 
angles  subtended  by  an   arc  at   all   points  of   the  cir 
cumference  are  equal. 

210.  The  angle  in  a  semicircle  is  a  right  angle. 

211.  If  AB  be   a  diameter  of  a  circle,  and  DC  a 
chord  at  right  angles   to  it,  \\\tn.ACBD  is  a  quadri 
lateral  of  which   AB  is  an  axis   of   symmetry.      The 
angles  BCA  and  ADB  being  each   a  right  angle,  the 
remaining  two   angles  DBC  and    CAD  are   together 
equal  to  a  straight  angle.      If  A'  and  B'  be  any  other 
points   on  the  arcs  DAC  and   CBD  respectively,  the 
/_CAD=-/_  CA'Dand  /_DBC=Z.DB'C,  and  ^CA'D 
-\-DB'C  =  a  straight  angle.   Therefore,  also,  /_B'CA' 
+  /  A'DB'  ==  a  straight  angle. 

Conversely,  if  a  quadrilateral  has  two  of  its  oppo 
site  angles  together  equal  to  two  right  angles,  it  is 
inscriptible  in  a  circle. 

*The  above  figures  and  proof  are  from  Beman  and  Smith's  New  Plane 
and  Solid  Geometry,  p.  129. 


io8 


GEOMETRIC  EXERCISES 


212.  The  angle  between  the  tangent  to  a  circle  and 
a  chord  which  passes  through  the  point  of  contact  is 
equal  to  the  angle  at  the  circumference  standing  upon 
that  chord  and  having  its  vertex  on  the  side  of  it  op 
posite  to  that  on  which  the  first  angle  lies. 

Let  AC  be  a  tangent  to  the  circle  at  A  and  AB  a 
chord.  Take  O  the  center  of  the  circle  and  draw  OA, 
OB.  Draw  OD  perpendicular  to  AB. 

Then  ^_BAC^=L  AOD  =  %  /  BOA. 


213.  Perpendiculars  to  diameters  at  their  extremi 
ties  touch  the  circle  at  these  extremities.  (See  Fig. 64). 
The  line  joining  the  center  and  the  point  of  intersection 


IN  PAPER  FOLDING  109 

of  two  tangents  bisects  the  angles  between  the  two 
tangents  and  between  the  two  radii.  It  also  bisects 
the  join  of  the  points  of  contact.  The  tangents  are 
equal. 

This  is  seen  by  folding  through  the  center  and 
the  point  of  intersection  of  the  tangents. 

Let  AC,  AB  be  two  tangents  and  ADEOF  the 
line  through  the  intersection  of  the  tangents  A  and 
the  center  O,  cutting  the  circle  in  D  and  F  and  BC 
in  E. 

Then  AC  or  AB  is  the  geometric  mean  of  AD  and 
AF\  AE  is  the  harmonic  mean;  andAO  the  arith 
metic  mean. 


AD-AF_  2AD-  AF 
~OA~    ~  AD-^AF' 

Similarly,  if  any  other  chord  through  A  be  ob 
tained  cutting  the  circle  in  P  and  R  and  BC  in  Q, 
then  AQ  is  the  harmonic  mean  and  AC  the  geometric 
mean  between  AP  and  AR. 


214.  Fold  a  right-angled  triangle  OCB  and  CA 
the  perpendicular  on  the  hypotenuse.  Take  D  in  AB 
such  that  OD=  OC  (Fig.  65). 


Then    O 

and       OA  :  OC=OC:  OB, 

OA  :  OD=OD\  OB. 


no  GEOMETRIC  EXERCISES 

A  circle  can  be  described  with  O  as  center  and 
OC  or  OD  as  radius. 

The  points  A  and  B  are  inverses  of  each  other 
with  reference  to  the  center  of  inversion  O  and  the 
circle  of  inversion  CDE. 


Fig.  65. 

Hence  when  the  center  is  taken  as  the  origin,  the 
foot  of  the  ordinate  of  a  point  on  a  circle  has  for  its 
mverse  the  point  of  intersection  of  the  tangent  and 
the  axis  taken. 

215.  Fold  FBG  perpendicular  to  OB.  Then  the 
line  FBG  is  called  the  polar  of  point  A  with  reference 
to  the  polar  circle  CDE  and  polar  center  O ;  and  A  is 
called  the  pole  of  FBG.  Conversely  B  is  the  pole  of 


2N  PAPER  FOLDING  Mi 

CA  and  CA   is  the  polar  of  B  with  reference  to  the 
same  circle. 

216.  Produce  OC  to  meet  FBG  in  F,  and  fold  AH 
perpendicular  to  OC. 

Then  F  and  Zf  are  inverse  points. 

AJfis  the  polar  of  7%  and  the  perpendicular  at  F 
to  <9^  is  the  polar  of  H. 


217.  The  points  A,  B,  F,  H,  are  concyclic. 

That  is,  two  points  and  their  inverses  are  con- 
cyclic  ;  and  conversely. 

Now  take  another  point  G  on  FBG.  Draw  OG, 
and  fold  AK  perpendicular  to  OG.  Then  K  and  G 
are  inverse  points  with  reference  to  the  circle  CDE. 

218.  The   points  F,  J3,  G  are  collinear,  while  their 
polars  pass  through  A. 

Hence,  the  polars  of  collinear  points  are  concur 
rent. 

219.  Points  so  situated  that  each  lies  on  the  polar 
of  the  other  are  called  conjugate  points,  and  lines  so 
related  that  each  passes  through  the  pole  of  the  other 
are  called  conjugate  lines. 

A  and  F  are  conjugate  points,  so  are  A  and  B,  A 
and  G. 

The  point  of  intersection  of  the  polars  of  two 
points  is  the  pole  of  the  join  of  the  points. 

220.  As  A  moves  towards  D,  B  also  moves  up  to  it. 
Finally  A  and  B  coincide  and  FBG  is  the  tangent  at  B. 


ii2  GEOMETRIC  EXERCISES 

Hence  the  polar  of  any  point  on  the  circle  is  the 
tangent  at  that  point. 

221.  As  A  moves  back  to  O,  B  moves  forward  to 
infinity.      The  polar  of  the  center  of  inversion  or  the 
polar  center  is  the  line  at  infinity. 

222.  The  angle  between  the  polars  of  two  points 
is  equal  to  the  angle  subtended  by  these  points  at  the 
polar  center. 

223.  The  circle  described  with  B  as  a  center  and 
BC  as  a  radius  cuts  the  circle  CDE  orthogonally. 

224.  Bisect  AB  in  L  and  fold  LN  perpendicular 
to  AB.      Then  all  circles  passing  through  A   and  B 
will  have  their  centers  on  this  line.     These  circles  cut 
the   circle    CDE    orthogonally.      The   circles   circum 
scribing    the   quadrilaterals   ABFH  and   ABGK  are 
such  circles.      AF  and  AG  are  diameters  of  the  re 
spective  circles.      Hence  if  two  circles  cut  orthogon 
ally  the  extremities  of  any  diameter  of  either  are  con 
jugate  points  with  respect  to  the  other. 

225.  The  points  O,  A,  //and  K  are  concyclic.    H, 
A,  K  being  inverses  of  points  on   the  line  FBG,  the 
inverse  of  a  line  is  a  circle   through  the  center  of  in 
version   and   the  pole   of  the  given  line,  these  points 
being  the  extremities  of  a  diameter ;   and  conversely. 

226.  If  DO  produced  cuts  the  circle  CDE  in  D', 
D  and  D'  are  harmonic  conjugates  of  A  and  B.     Sim- 


IN  PAPER  FOLDING  11.3 

ilarly,  if  any  line  through  B  cuts  AC  in  A'  and  the 
circle  CDE  in  d  and  dk  ',  then  ^/  and  */'  are  harmonic 
conjugates  of  A'  and  B. 

227.  Fold  any  line  LM=LB  =  LA,  and  J/6>'  per 
pendicular  to  LM  meeting  AB  produced  in  O'. 

Then  the  circle  described  with  center  O'  and  ra 
dius  O'Mcuts  orthogonally  the  circle  described  with 
center  L  and  radius  LM. 
Now  OL2  =  O£*  +  L£2, 

and  <9'Z2  =  O'M*      ZJ/2. 


.-.  ZTVis  the  radical  axis  of  the  circles  O  (OC} 
and  O'(O'M). 

By  taking  other  points  in  the  semicircle  AMB  and 
repeating  the  same  construction  as  above,  we  get  two 
infinite  systems  of  circles  co-axial  with  O(OC}  and 
O\O'M),  viz.,  one  system  on  each  side  of  the  radical 
axis,  LN.  The  point  circle  of  each  system  is  a  point, 
A  or  B,  which  may  be  regarded  as  an  infinitely  small 
circle. 

The  two  infinite  systems  of  circles  are  to  be  re 
garded  as  one  co-axial  system,  the  circles  of  which 
range  from  infinitely  large  to  infinitely  small  —  the 
radical  .axis  being  the  infinitely  large  circle,  and  the 
limiting  points  the  infinitely  small.  This  system  of 
co-axial  circles  is  called  the  limiting  point  species. 

If  two  circles  cut  each  other  their  common  chord 
is  their  radical  axis.  Therefore  all  circles  passing 


ii4  GEOMETRIC  EXERCISES 

through  A   and  B  are  co-axial.      This   system    of  co 
axial  circles  is  called  the  common  point  species. 

228.  Take  two  lines  OAB  and  OPQ.      From   two 
points  A  and  B  in  OAB  draw  AP,  BQ  perpendicular 
to    OPQ.      Then  circles  described  with  A   and   B  as 
centers  and  AP  and  BQ  as  radii  will  touch  the  line 
OPQ  at  P  and  Q. 

Then  OA-.OB  =  AP:BQ. 

This  holds  whether  the  perpendiculars  are  towards 
the  same  or  opposite  parts.  The  tangent  is  in  one 
case  direct,  and  in  the  other  transverse. 

In  the  first  case,  O  is  outside  AB,  and  in  the  sec 
ond  it  is  between  A  and  B.  In  the  former  it  is  called 
the  external  center  of  similitude  and  in  the  latter  the 
internal  centre  of  similitude  of  the  two  circles. 

229.  The  line  joining   the  extremities  of  two  par 
allel  radii  of  the  two  circles  passes  through   their  ex 
ternal  center  of  similitude,  if  the  radii  are  in  the  same 
direction,  and  through  their  internal  center,  if   they 
are  drawn  in  opposite  directions. 

230.  The  two  radii  of  one  circle  drawn  to  its  points 
of  intersection  with  any  line  passing  through  either 
center  of   similitude,  are  respectively  parallel    to  the 
two  radii  of  the  other  circle  drawn  to  its  intersections 
with  the  same  line. 

231.  All  secants  passing  through  a  center  of  simil 
itude  of  two  circles  are  cut  in  the  same  ratio  by  the 
circles. 


IN  PA  PER  FOLDING  1  1  5 

232..  If  B\>  D\t  and  B^  D<t  be  the  points  of  inter 
section,  B\,  £2,  and  D\,  DI  being  corresponding 
points, 

=  ODl  -  OBi  = 


Hence  the  inverse  of  a  circle,  not  through  the  cen 
ter  of  inversion  is  a  circle. 


Fig.  66. 

The  center  of  inversion  is  the  center  of  similitude 
of  the  original  circle  and  its  inverse. 

The  original  circle,  its  inverse,  and  the  circle  of 
inversion  are  co-axial. 

233.  The  method  of  inversion  is  one  of  the  most 
important  in  the  range  of  Geometry.      It  was  discov 
ered  jointly  by  Doctors  Stubbs  and   Ingram,  Fellows 
of  Trinity  College,  Dublin,  about  1842.      It  was  em 
ployed  by  Sir  William  Thomson  in  giving  geometric 
proof  of  some  of  the  most  difficult  propositions  in  the 
mathematical  theory  of  electricity. 

SECTION  II.— THE  PARABOLA. 

234.  A   parabola  is  the   curve   traced   by  a  point 
which  moves  in  a  plane  in  such  a  manner  that  its  dis- 


n6 


GEOMETRIC  EXERCISES 


tance  from  a  given   point  is  always  equal  to  its  dis 
tance  from  a  given  straight  line. 

235.  Fig.  67  shows  how  a  parabola  can  be  marked 
on  paper.  The  edge  of  the  square  MN  is  the  direct 
rix,  O  the  vertex,  and  F  the  focus.  Fold  through  OX 
and  obtain  the  axis.  Divide  the  upper  half  of  the 


Fig.  67. 

square  into  a  number  of  sections  by  lines  parallel  to 
the  axis.  These  lines  meet  the  directrix  in  a  number 
of  points.  Fold  by  laying  each  of  these  points  on  the 
focus  and  mark  the  point  where  the  corresponding 
horizontal  line  is  cut.  The  points  thus  obtained  lie 
on  a  parabola.  The  folding  gives  also  the  tangent  to 
the  curve  at  the  point. 


IN  PAPER  FOLDING 


117 


236.  FL  which  is  at  right  angles  to  OX  is  called 
the  semi-latus  rectum. 

237.  When  points  on  the  upper  half  of  the   curve 
have  been  obtained,  corresponding  points  on  the  lower 
half  are  obtained  by  doubling  the  paper  on  the  axis 
and  pricking  through  them. 

238.  When  the  axis  and  the  tangent  at  the  vertex 
are  taken  as  the  axes  of  co-ordinates,  and  the  vertex 
as  origin,  the  equation  of  the  parabola  becomes 

j>2  =  4:ax  or 

Y 


The  parabola  may  be  denned  as  the  curve  traced 
by  a  point  which  moves  in  one  plane  in  such  a  manner 
that  the  square  of  its  distance  from  a  given  straight 
line  varies  as  its  distance  from  another  straight  line ; 
or  the  ordinate  is  the  mean  proportional  between  the 


1  1  8  GEOME  TRIG  EXERCISES 

abscissa,  and  the  latus  rectum  which  is  equal  to  4-  OF. 
Hence  the  following  construction. 

Take  O  T  in  FO  produced  =  4  •  OF. 

Bisect  TN'm  M. 

Take  Q  in  OYsuch  that  MQ  =  MN=MT. 

Fold  through  Q  so  that  QP  may  be  at  right  angles 
to  OY. 

Let  P  be  the  point  where  QP  meets  the  ordinate 
of  N. 

Then  P  is  a  point  on  the  curve. 

239.  The  subnormal^  2  OF  and  FP—FG  =  FT'. 
These  properties  suggest  the  following  construc 
tion. 

Take  ^Vany  point  on  the  axis. 

On   the   side   of  N  remote   from   the   vertex   take 


Fold  NP  perpendicular  to  OG  and  find  P  in  NP 
such  that  FP  =  FG. 

Then  P  is  a  point  on  the  curve. 

A  circle  can  be  described  with  F  as  center  and  FG, 
FP  and  FT'  as  radii. 

The  double  ordinate  of  the  circle  is  also  the  double 
ordinate  of  the  parabola,  i.  e.,  P  describes  a  parabola 
as  Amoves  along  the  axis. 

240.  Take  any  point  N'  between  O  and  F  (Fig.  69). 
Fold  RN'P  at  right  angles  to  OF. 
Take  R  so  that  OR  =  OF. 
Fold  ^/^perpendicular  to  OR,  N  being  on  the  axis. 


IN  PAPER  FOLDING 


119 


Fold  NP  perpendicular  to  the  axis. 
Now,  in  OX  take  OT=OW. 
Take  P  in  RN'  so  that  FP  =  FT. 
Fold  through  P'F  cutting  NP  in  P. 
Then  P  and  .P  are  points  on  the  curv 


N'    /F          N 


Fig.  69. 

241.  Arand  N'  coincide  when   PFP   is  the  latus 
rectum. 

As  N'  recedes  from  F  to  O,  7^  moves  forward  from 

infinity. 

At  the  same  time,  Amoves  toward  O,  and  T'(OT'= 
moves  in  the  opposite  direction  toward  infinity. 

242.  To  find  the  area  of  a   parabola  bounded  by 
the  axis  and  an  ordinate. 

Complete  the  rectangle  ONPK.     Let  OK  be  di- 


\20  GE  OME  TRIG  EXER  CISE  S 

vided  into  n  equal  portions  of  which  suppose  Om  to 
contain  r  and  mn  to  be  the  (r  -f-  1)'*.  Draw  mp,  nq  at 
right  angles  to  OK  meeting  the  curve  in  p,  q,  and  pn' 
at  right  angles  to  nq.  The  curvilinear  area  OPK  is 
the  limit  of  the  sum  of  the  series  of  rectangles  con 
structed  as  mn'  on  the  portions  corresponding  to  mn. 

But  mi/»  :  I — I  NK—pm  •  mn  :  PK-  OK, 
and,  by  the  properties  of  the  parabola, 
pm\PK=Om*\  OK* 

and  mn  :  OK=  1  :  n. 

.  • .  pm  •  mn\PK'  OK=  r*  :  ;z3. 
rl 

.  • .  i—±pn  =     x  en  A^- 
w 

Hence  the  sum  of  the  series  of  rectangles 
p  _|_  22 -f- 32 +  («—  I)2 


(«  — 1)«(2«— 1) 


1-2-3-w3 


=  -J  of  nn^VA'in  the  limit,  i.  e.,  when  «  is  oo. 
.  •.  The  curvilinear  area  OPK=^  of  cZi^VA',  and  the 
parabolic  area  ^^TV^^f  of  I  —  \NK. 


243.  The  same  line  of  proof  applies  when  any 
diameter  and  an  ordinate  are  taken  as  the  boundaries 
of  the  parabolic  area. 


IN  PAPER  FOLDING 


121 


SECTION  III.— THE  ELLIPSE. 

244.  An  ellipse  is  the  curve  traced  by  a  point  which 
moves  in  a  plane  in  such  a  manner  that  its  distance 
from  a  given  point  is  in  a  constant  ratio  of  less  in 
equality  to  its  distance  from  a  given  straight  line. 

Let  Fbe  the  focus,  OYthe  directrix,  and  XX'  the 
perpendicular  to  O  Y  through  F.  Let  FA  \AObe  the 


Fig.  70. 

constant  ratio,  FA  being  less  than  AO.     A  is  a  point 
on  the  curve  called  the  vertex. 

As  in  §  116,  find  A'  in  XX'  such  that 
FA':A'O  =  FA  :AO. 

Then  A'  is  another  point  on  the  curve,  being  a 
second  vertex. 

Double  the  line  AA'  on  itself  and  obtain  its  middl'e 
point  C,  called  the  center,  and  mark  F'  and  O'  corre 
sponding  to  F  and  O.  Fold  through  O'  so  that  O' Y1 


122  GEOMETRIC  EXERCISES 

may  be  at  right  angles  to  XX' .     Then  F*  is  the  sec 
ond  focus  and  O'Y'  the  second  directrix. 

By  folding  A  A ',  obtain  the  perpendicular  through  C. 

FA  :AO  =  FA'  -.  A' O 

=  FA-\-FA':AO  +  A'O 
=  AA':  OO' 
=  CA  :  CO. 

Take  points  B  and  B'  in  the  perpendicular  through 
C  and  on  opposite  sides  of  it,  such  that  FB  and  FB' 
are  each  equal  to  CA.  Then  B  and  B'  are  points  on 
the  curve. 

AA'  is  called  the  major  axis,  and  BB'  the  minor 
axis. 

245.  To  find  other  points  on  the  curve,  take  any 
point  E  in  the  directrix,  and  fold  through  E  and  A, 
and  through  E  and  A '.  Fold  again  through  E  and  F 
and  mark  the  point  P  where  FA'  cuts  EA  produced. 
Fold  through  PF  and  P  on  EA' .  Then  P  and  P  are 
points  on  the  curve. 

Fold  through  P  and  P'  so  that  KPL  and  K' L' P' 
are  perpendicular  to  the  directrix,  K  and  K'  being  on 
the  directrix  and  L  and  L'  on  EL. 

FL  bisects  the  angle  A'FP, 

.-.   tLFP  =  LPLF  and 
FP\PK=PL\PK 
=  FA  :  AO. 
And 

FP  -.PK'=P'L'  -.PK' 


IN  PAPER  FOLDING  123 


=  FA  :AO. 

If  EO  =  FO,  FP  is  at  right  angles  to  FO,  and 
FP=FP'.  PP'  is  the  latus  rectum. 

246.  When  a  number  of  points  on  the  left  half  of 
the  curve  are  found,  corresponding  points  on  the  other 
half   can   be   marked  by  doubling   the  paper  on   the 
minor  axis  and  pricking  through  them. 

247.  An  ellipse  may  also  be  defined  as  follows  : 

If  a  point  P  move  in  such  a  manner  that  /W2 
\AN-NA'  is  a  constant  ratio,  PN  being  the  distance 
of  P  from  the  line  joining  two  fixed  points  A,  A  ',  and 
N  being  between  A  and  A',  the  locus  of  P  is  an  ellipse 
of  which  AA'  is  an  axis. 


248.  In  the  circle, 

In  the  ellipse  PN'2  :  AN-NA'  is  a  constant  ratio. 

This  ratio  may  be  less  or  greater  than  unity.  In 
the  former  case  £APA'  is  obtuse,  and  the  curve  lies 
within  the  auxiliary  circle  described  on  AA'  as  diam 
eter.  In  the  latter  case,  /  APA'  is  acute  and  the  curve 
is  outside  the  circle.  In  the  first  case  AA'  is  the 
major,  and  in  the  second  it  is  the  minor  axis. 

249.  The  above  definition  corresponds  to  the  equa 
tion 


when  the  vertex  is  the  origin. 


124 


GEOMETRIC  EXERCISES 


250.  AN*  NA'  is  equal  to  the  square  on  the  ordi- 
nate    QN   of    the    auxiliary    circle,    and    PN ':  QN  = 
BC-.AC. 

251.  Fig.  71  shows  how  the  points  can  be  deter 
mined   when   the   constant   ratio   is   less   than    unity. 
Thus,  lay  off  CD= -AC,  the  semi-major  axis.   Through 
E  any  point  of  ^Cdraw  DE  and  produce  it  to  meet 

Q 


the  auxiliary  circle  in  Q.  Draw  JB' E  and  produce  it 
to  meet  the  ordinate  QN  in  P.  Then  is  PN:  QN 
—  &C\DC=BC\AC.  The  same  process  is  appli 
cable  when  the  ratio  is  greater  than  unity.  When 
points  in  one  quadrant  are  found,  corresponding  points 
in  other  quadrants  can  be  easily  marked. 

252.   If  P  and  P'  are  the  extremities  of  two  conju 
gate  diameters  of  an  ellipse  and  the  ordinates   MP 


IN  PAPER  FOLDING 


and  M' P'  meet  the  auxiliary  circle  in  Q  and  Q',  the 
angle  QCQ'  is  a  right  angle. 

Now  take  a  rectangular  piece  of  card  or  paper  and 
mark  on  two  adjacent  edges  beginning  with  the  com 
mon  corner  lengths  equal  to  the  minor  and  major 
axes.  By  turning  the  card  round  C  mark  correspond 
ing  points  on  the  outer  and  inner  auxiliary  circles. 
Let  Q,  R  and  Q',  Rr  be  the  points  in  one  position. 
Fold  the  ordinates  QM  and  Q'M',  and  RP  and  R'P't 
perpendiculars  to  the  ordinates.  Then  P  and  P'  are 
points  on  the  curve. 


Fig.  72. 

253.  Points  on  the  curve  may  also  be  easily  deter 
mined  by  the  application  of  the  following  property  of 
the  conic  sections. 

The  focal  distance  of  a  point  on  a  conic  is  equal 


126 


GEOMETRIC  EXERCISES 


to  the  length  of  the  ordinate  produced   to  meet  the 
tangent  at  the  end  of  the  latus  rectum. 

254.  Let  A  and  A'  be  any  two  points.  Draw  A  A' 
and  produce  the  line  both  ways.  From  any  point  D 
in  A' A  produced  draw  DR  perpendicular  to  AD.  Take 
any  point  R  in  DR  and  draw  RA  and  RA '.  Fold  AP 
perpendicular  to  AR,  meeting  RA'  in  P.  For  different 
positions  of  R  in  DR,  the  locus  of  P  is  an  ellipse,  of 
which  AA'  is  the  major  axis. 


Fig.  73- 

Fold  PN  perpendicular  to  A  A'. 
Now,  because  PN'is  parallel  to  RD, 

PN:A'N=RD:A'D. 
Again,  from  the  triangles,  APN and  DAR, 

PN\AN=AD\  RD. 

.-.  PN*'.AN'A'N=AD\A'D,  a  constant  ratio, 
less  than  unity,  and  it  is  evident  from  the  construc 
tion  that  jVmust  lie  between  A  and  A'. 

SECTION  IV.— THE  HYPERBOLA. 

255.  An  hyperbola  is  the  curve  traced  by  a  point 
which  moves  in  a  plane  in  such  a  manner  that  its 


IN  PAPER  FOLDING.  127 

distance  from  a  given  point  is  in  a  constant  ratio  of 
greater  inequality  to  its  distance  from  a  given  straight 
line. 

256.  The  construction  is  the  same  as  for  the  el 
lipse,  but  the  position  of  the  parts  is  different.     As 
explained  in  §  119,  X,  A'  lies  on  the  left  side  of  the 
directrix.     Each  directrix  lies  between  A  and  A',  and 
the  foci   lie  without   these    points.     The  curve  con 
sists  of  two  branches  which  are  open  on   one  side. 
The  branches  lie  entirely  within  two  vertical  angles 
formed  by  two  straight  lines  passing  through  the  cen 
ter  which  are  called  the  asymptotes.      These  are  tan 
gents  to  the  curve  at  infinity. 

257.  The  hyperbola  can  be  defined  thus  :   If  a  point 
P  move  in  such  a  manner  that  PN^  :  AN  -  NA'  is  a 
constant  ratio,  PN  being  the  distance  of  P  from  the 
line  joining  two  fixed  points  A  and  A',  and  TV  not 
being  between  A  and  A',  the  locus  of  P  is  an  hyper 
bola,  of  which  AA'  is  the  transverse  axis. 

This  corresponds  to  the  equation 


where  the  origin  is  at  the  right-hand  vertex  of  the 
hyperbola. 

Fig.  74  shows  how  points  on  the  curve  may  be 
found  by  the  application  of  this  formula. 

Let  C  be  the  center  and  A  the  vertex  of  the  curve. 


GEOMETRIC  EXERCISES 


12  S 


CA'  =  CA  =  CA'  =  a. 

Fold  CD  any  line  through  C  and  make  C£>  =  CA. 
Fold  DN  perpendicular  to  CD.  Fold  NQ  perpen 
dicular  to  CA  and  make  NQ  =  DN.  Fold  Q  A"  cut 
ting  CA  in  S.  Fold  ^'S  cutting  (Win  P. 


Fig.  74- 

Then  />  is  a  point  on  the  curve. 
For,    since  DN  is    tangent   to  the   circle   on   the 
diameter  A' A 

DN*  =  AN-  (2CA  +  AN), 
or  since  QN=DN, 


IN  PAPER  FOLDING.  za? 

QN      A"C 


Squaring,  — ^—  _  =  _ , 

or  y2=—2(2ax  +  x2). 

If  QN=b  then  ^  is  the  focus  and  CD  is  one  of 
the  asymptotes.  If  we  complete  the  rectangle  on 
AC  and  BC  the  asymptote  is  a  diagonal  of  the  rect 
angle. 

258.  The  hyperbola  can  also  be  described  by  the 
property  referred  to  in  §  253. 

259.  An  hyperbola  is   said  to  be  equilateral  when 
the  transverse  and  conjugate  axes  are  equal.      Here 
a  =  fr,  and  the  equation  becomes 

In  this  case  the  construction  is  simpler  as  the  ordi- 
nate  of  the  hyperbola  is  itself  the  geometric  mean  be 
tween  AN  and  A'N,  and  is  therefore  equal  to  the  tan 
gent  from  TVto  the  circle  described  on  A  A'  as  diameter. 

260.  The  polar  equation  to  the  rectangular  hyper 
bola,  when  the  center  is  the  origin  and  one  of  the  axes 
the  initial  line,  is 

r2  cos  26  =  a2 

or  r2  = -pj  •  a. 

cos26 

Let  OX,  OYbe  the  axes;  divide  the  right  angle 
VOX  into  a  number  of  equal  parts.  Let  XOA,  A  OB 


130 


GEOMETRIC  EXERCISES 


be  two  of  the  equal  angles.  Fold  XB  at  right  angles 
to  OX.  Produce  BO  and  take  OF=  OX.  Fold  OG 
perpendicular  to  BF  and  find  G  in  OG  such  that  FGB 
is  a  right  angle.  Take  OA  =  OG.  Then  A  is  a  point 
on  the  curve. 


Fig.  75- 

Now,  the  angles  XOA  and  AOB  being  each  0, 
OB= 


COS 


26' 


And  O42  =OG^= 


a 
COS29**" 


261.  The  points  of  trisection  of  a  series  of  conter 
minous  circular  arcs  lie  on  branches  of  two  hyperbolas 
of  which  the  eccentricity  is  2.  This  theorem  affords 
a  means  of  trisecting  an  angle.* 

*  See  Taylor's  Ancient  and  Modern  Geometry  of  Conies,  examples  308,  390 
with  footnote. 


XIV.   MISCELLANEOUS  CURVES. 

262.  I  propose  in  this,   the  last  chapter,  to  give 
hints  for  tracing  certain  wall-known  curves. 

THE  CISSOID.* 

263.  This  word  means  ivy-shaped  curve.     It  is  de 
nned  as  follows:   Let  OQA  (Fig.  76)  be  a  semicircle 
on  the  fixed  diameter  OA,  and  let  QM,  RN  be  two 
ordinates  of  the  semicircle   equidistant  from  the  cen 
ter.      Draw  OR  cutting   QM  in  P.     Then  the  locus 
of  P  is  the  cissoid. 

If  OA=2a,  the  equation  to  the  curve  is 

/(2a  —  x)=x*. 

Now,  let  PR  cut  the  perpendicular  from  C  in  D 
and  draw  AP  cutting  CD  in  E. 

RN:CD  =  ON:  OC=AM~AC=PM:EC, 
.-.   RN-,PM=CD:CE. 
But  RN\  PM=ON\  OM=ON:  AN^ON*  :  NR* 


If  CF  be  the  geometric  mean  between  CD  and  CE, 

*See  Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of  Ele 
mentary  Geometry,  p.  44. 


GEOMETRIC  EXERCISES 
CD:CF=OC:CD 


132 


.  • .    CD  and  CF  are  the  two  geometric  means  be 
tween  OC  and  CE. 


M  C  N  F  A 

Fig.  76. 

264.  The  cissoid  was  invented  by  Diocles  (second 
century  B.  C.)  to  find  two  geometric  means  between 
two  lines  in   the  manner  described  above.      OC  and 
CE  being  given,  the  point  P  was  determined  by  the 
aid  of  the  curve,  and  hence  the  point  D. 

265.  If  PD  and  DR  are  each  equal  to  OQ,  then 
the  angle  AOQ  is  trisected  by  OP. 


IN  PAPER  FOLDING.  133 

Draw  QR.     Then  QR  is  parallel  to  OA,  and 


THE  CONCHOID  OR  MUSSEL-SHAPED  CURVE.* 


266.  This  curve  was  invented  by 
150  B.  C.).  Let  O  be  a 
fixed  point,  a  its  dis- 
tance  from  a  fixed  line, 
DM,  and  let  a  pencil  of 
rays  through  O  cut  DM. 
On  each  of  these  rays 
lay  off,  each  way  from  its 
intersection  with  DM,  a 
segment  b.  The  locus 
of  the  points  thus  deter 
mined  is  the  conchoid. 
According  as  b  >,  =, 
or  <#,  the  origin  is  a 
node,  a  cusp,  or  a  con 
jugate  point.  The  fig- 
ure|  represents  the  case 
when  b  >  a. 


Nicomedes  (c. 


Fig.  77- 


267.  This  curve  also 
was  employed  for  finding 
two  geometric  means,  and  for  the  trisection  of  an  angle. 


*See  Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of  Elt- 
mentary  Geometry,  p.  40. 

tFrom  Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of 
Elementary  Geometry,  p.  46. 


134 


GEOMETRIC  EXERCISES 


Let  OA  be  the  longer  of  the  two  lines  of  which 
two  geometric  means  are  required. 

Bisect  OA  in  B\  with  O  as  a  center  and  OB  as  a 
radius  describe  a  circle.  Place  a  chord  BC  in  the 
circle  equal  to  the  shorter  of  the  given  lines.  Draw 
AC  and  produce  AC  and  BC  to  D  and  E,  two  points 
collinear  with  O  and  such  that  DE  —  OB,  or  BA. 


Fig.  78. 

Then  ED  and  CE  are  the  two  mean  proportionals 
required. 

Let  OE  cut  the  circles  in  E  and  G. 
By  Menelaus's  Theorem,* 

BC-ED  •  OA=CE  -  OD  -BA 
...  BC'OA=CE-OD 


BC  _  OD 
~CE  ~~  OA 
BE  OD  +  OA  GE 


'  CE  ~  OA  OA' 

»See  Beman  and  Smith's  New  Plane  and  Solid  Geometry,  p.  240. 


IN  PAPER  FOLDING.  135 


But  GE  •  EF=  BE  -EC. 

.-.    GE  -OD  = 
.-.    OA  -OD  = 


The  position  of  E  is  found  by  the  aid  of  the  con 
choid  of  which  AD  is  the  asymptote,  O  the  focus,  and 
DE  the  constant  intercept. 

268.  The  trisection  of  the  angle  is  thus  effected. 
In  Fig.  77,  let  <£  =  /  MOV,  the  angle  to  be  trisected. 
On  OM  lay  off  OM=b,  any  arbitrary  length.     With 
M  as  a  center  and  a  radius  b  describe  a  circle,  and 
through  M  perpendicular  to  the  axis  of  X  with  origin 
O  draw  a  vertical  line  representing  the  asymptote  of 
the  conchoid  to  be  constructed.     Construct  the  con 
choid.   Connect  O  with  A,  the  intersection  of  the  circle 
and  the  conchoid.     Then  is  /  A  OY  one  third  of  <p.* 

THE  WITCH. 

269.  If  OQA  (Fig.  79)  be  a  semicircle  and  NQ  an 
ordinate  of  it,  and  NP  be  taken  a  fourth  proportional 
to  ON,  OA  and  (M7,  then  the  locus  of  P  is  the  witch. 

Fold  AM  at  right  angles  to  OA. 
Fold  through  O,  Q,  and  M. 
Complete  the  rectangle  NAMP. 
PN\  QN=OM:  OQ 
=  OA\ON. 

"-Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of  Elemen 
tary  Geometry,  p.  46. 


136  GEOMETRIC  EXERCISES 

Therefore  P  is  a  point  on  the  curve. 
Its  equation  is, 


Fig-  79- 

This  curve  was  proposed  by  a  lady,  Maria  Gaetana 
Agnesi,  Professor  of  Mathematics  at  Bologna. 

THE  CUBICAL  PARABOLA. 

270.   The  equation  to  this  curve  is  a*y  =  x*. 

Let  OX  and  OYbe  the  rectangular  axes,  OA=a, 
and  OX=x. 

In  the  axis  OY  take  OB  =  x. 

Draw  BA  and  draw  AC  at  right  angles  to  AB  cut 
ting  the  axis  O Kin  C. 


IN  PAPER  FOLDING. 


137 


Draw  CX,  and  draw  XYat  right  angles  to  CX. 
Complete  the  rectangle  XOY. 
P  is  a  point  on  the  curve. 


Fig.  80. 


THE  HARMONIC  CURVE  OR  CURVE  OF  SINES. 

271.  This  is  the  curve  in  which  a  musical  string 
vibrates  when  sounded.  The  ordinates  are  propor 
tional  to  the  sines  of  angles  which  are  the  same  frac 
tions  of  four  right  angles  that  the  corresponding  ab 
scissas  are  of  some  given  length. 

Let  AB  (Fig.  81)  be  the  given  length.   Produce  BA 


i38 


GEOMETRIC  EXERCISES 


to  C  and  fold  AD  perpendicular  to  AB.  Divide  the 
right  angle  DAC  into  a  number  of  equal  parts,  say, 
four.  Mark  on  each  radius  a  length  equal  to  the  am 
plitude  of  the  vibration,  AC—AP=AQ  =  AR  =  AD. 

From  points/1,  Q,  R  fold  perpendiculars  to  A  C; 
then  PP',  QQ',  RR',  and  DA  are  proportional  to  the 
sines  of  the  angles  PAC,  QAC,  RAC,  DAC. 

Now,  bisect  AB  in  E  and  divide  AE  and  EB  into 
twice  the  number  of  equal  parts  chosen  for  the  right 


C  ?'  Q' 


S'  T'  U'  V 


angle.  Draw  the  successive  ordinates  SS',  TT',  UU\ 
VV,  etc.,  equal  to  PP,  QQ',  RR',  DA,  etc.  Then 
S,  T,  U,  V  are  points  on  the  curve,  and  V  is  the 
highest  point  on  it.  By  folding  on  VV  and  pricking 
through  S,  T,  U,  V,  we  get  corresponding  points  on 
the  portion  of  the  curve  VE.  The  portion  of  the 
curve  corresponding  to  EB  is  equal  to  A  VE  but  lies 
on  the  opposite  side  of  AB.  The  length  from  A  to  E 
is  half  a  wave  length,  which  will  be  repeated  from  E 


PAPER  FOLDING 


139 


to  B  on  the  other  side  of  AB.  E  is  a  point  of  inflec 
tion  on  the  curve,  the  radius  of  curvature  there  be 
coming  infinite. 


THE  OVALS  OF  CASSINI. 


272.  When  a  point  moves  in  a  plane  so  that  the 
product  of  its  distances  from  two  fixed  points  in  the 
plane  is  constant,  it  traces  out  one  of  Cassini's  ovals. 
The  fixed  points  are  called  the  foci.  The  equation  of 


X         MA 


A' 


B' 
Fig.  82. 

the  curve  is  rr' =k?,  where  r  and  r  are  the  distances 
of  any  point  on  the  curve  from  the  foci  and  k  is  a  con 
stant. 

Let  F  and  F'  be  the  foci.  Fold  through  F  and 
jF".  Bisect  FF'  in  C,  and  fold  BCB'  perpendicular  to 
FF'.  Find  points  B  and  B'  such  that  FB  and  FBf 
are  each  =k.  Then  B  and  B'  are  evidently  points  on 
the  curve. 


1  40  GEOMETRIC  EXERCISES 

Fold  FK  perpendicular  to  FF'  and  make  FK=k, 
and  on  FF'  take  CA  and  CA'  each  equal  to  CK.  Then 
A  and  A'  are  points  on  the  curve. 

For  CA*  =CK*  =  CF*  -f 


Produce  FA  and  take  AT=FK.  In  A  T  take  a 
point  J/and  draw  MK.  Fold  A'^/'  perpendicular  to 
MK  meeting  FA'  in  J/'. 

Then  FM-FM'=&. 

With  the  center  F  and  radius  FM,  and  with  the 
center  F'  and  radius  FM',  describe  two  arcs  cutting 
each  other  in  P.  Then  P  is  a  point  on  the  curve. 

When  a  number  of  points  between  A  and  B  are 
found,  corresponding  points  in  the  other  quadrants 
can  be  marked  by  paper  folding. 

When  FF'=V*k  and  rr  =  \k*  the  curve  as 
sumes  the  form  of  a  lemniscate.  (§  279.) 

When  FF'  is  greater  than  V  2k,  the  curve  consists 
of  two  distinct  ovals,  one  about  each  focus. 

THE  LOGARITHMIC  CURVE. 

273.   The  equation  to  this  curve  \sy  =  a*. 

The  ordinate  at  the  origin  is  unity. 

If  the  abscissa  increases  arithmetically,  the  ordi 
nate  increases  geometrically. 

The  values  of  y  for  integral  values  of  x  can  be  ob 
tained  by  the  process  given  in  §  108. 


IN  PAPER  FOLDING.  141 

The  curve  extends  to  infinity  in  the  angular  space 
XOY. 

If  x  be  negative  y=  —  and  approaches  zero  as  x 
ax 

increases  numerically.      The  negative  side  of  the  axis 
OX  is  therefore  an  asymptote  to  the  curve. 

THE  COMMON  CATENARY. 

274.  The  catenary  is  the  form  assumed  by  a  heavy 
inextensible  string  freely  suspended  from  two  points 
and  hanging  under  the  action  of  gravity. 

The  equation  of  the  curve  is 


the  axis  of  y  being  a  vertical  line  through  the  lowest 
point  of  the  curve,  and  the  axis  of  x  a  horizontal  line 
in  the  plane  of  the  string  at  a  distance  c  below  the 
lowest  point  ;  c  is  the  parameter  of  the  curve,  and  e 
the  base  of  the  natural  system  of  logarithms. 

When  x  =  c       =C-e^- 


when  x  —  2c,  y  =  —  (fi  -f-  e~2~)  and  so  on, 


275.  From  the  equation 

,=£(<*-»-,- 

e  can  be  determined  graphically. 


I42 


GEOMETRIC  EXERCISES 


I/}-2  —  f2  is  found  by  taking  the  geometric  mean  be 
tween  y  -j-  c  and  jv  —  r. 

THE  CARDIOID  OR  HEART-SHAPED  CURVE. 

276.  From  a  fixed  point  O  on  a  circle  of  radius  a 
draw  a  pencil  of  lines  and  take  off  on  each  ray,  meas 
ured  both  ways  from  the  circumference,  a  segment 
equal  to  2a.  The  ends  of  these  lines  lie  on  a  cardioid. 


Fig.  83. 


The  equation  to  the  curve  is  r  =  0(1  -f  cos  #)• 
The  origin  is  a  cusp  on  the  curve.      The  cardioid 

is  the  inverse  of  the  parabola  with   reference  to  its 

focus  as  center  of  inversion. 


THE  LIMACON. 

277.  From  a  fixed  point  on  a  circle,  draw  a  num 
ber  of  chords,  and  take  off  a  constant  length  on  each 
of  these  lines  measured  both  ways  from  the  circum 
ference  of  the  circle. 


IN  PAPER  FOLDING  143 

If  the  constant  length  is  equal  to  the  diameter  of 
the  circle,  the  curve  is  a  cardioid. 

If  it  be  greater  than  the  diameter,  the  curve  is 
altogether  outside  the  circle. 

If  it  be  less  than  the  diameter,  a  portion  of  the 
curve  lies  inside  the  circle  in  the  form  of  a  loop. 

If  the  constant  length  is  exactly  half  the  diameter, 
the  curve  is  called  the  trisectrix,  since  by  its  aid  any 
angle  can  be  trisected. 

The  equation  is  r  =  acos6-\-  b. 


The  first  sort  of  liir^on  is  the  inverse  of  an  ellipse ; 
and  the  second  sort  is  the  inverse  of  an  hyperbola, 
with  reference  to  a  focus  as  a  center.  The  loop  is  the 
inverse  of  the  branch  about  the  other  focus. 

278.  The  trisectrix  is  applied  as  follows : 
Let  AOB  be  the  given  angle.   Take  OA,  OB  equal 
to  the  radius  of  the  circle.     Describe  a  circle  with  the 
center   O  and  radius   OA  or   OB.      Produce  AO  in- 


i44  GEOMETRIC  EXERCISES 

definitely  beyond  the  circle.  Apply  the  trisectrix  so 
that  O  may  correspond  to  the  center  of  the  circle  and 
OB  the  axis  of  the  loop.  Let  the  outer  curve  cut  AO 
produced  in  C.  Draw  BC  cutting  the  circle  in  D, 
Draw  OD. 


Fig.  85. 

Then  ^ACB  is  \  of 
For  CD 

/_CJ3O 
/_ODB 


THE  LEMNISCATE  OF  BERNOULLI. 

279.  The  polar  equation  to  the  curve  is 

r2  =  a2cos26. 

Let  O  be  the  origin,  and  OA=a. 
Produce  AO,  and  draw  OD  at  right  angles  to  OA 
Take  the  angle  A  OP  =8  and  A  OB  '=--26. 
Draw  AB  perpendicular  to  OB. 
In  AO  produced  take  OC=^OB. 


IN  PAPER  FOLDING 

Find  D  in  OD  such  that  CD  A  is  a  right  angle. 

Take  OP  =  OD. 

P  is  a  point  on  the  curve. 


145 


=  OB-OA 


=  a2  cos  2  6. 

As  stated  above,  this  curve  is  a  particular  case  of 
the  ovals  of  Cassini. 


Fie.  86. 


It  is  the  inverse  of  the  rectangular  hyperbola,  with 
reference  to  its  center  as  center  of  inversion,  and  also 
its  pedal  with  respect  to  the  center. 

The  area  of  the  curve  is  a1. 


THE  CYCLOID. 

280.  The  cycloid  is  the  path  described  by  a  point 
on  the  circumference  of  a  circle  which  is  supposed  to 
roll  upon  a  fixed  straight  line. 

Let  A  and  A'  be  the  positions  of  the  generating 
point  when  in  contact  with  the  fixed  line  after  one 


1 46  GEOMETRIC  EXERCISES 

complete  revolution  of  the  circle.  Then  A  A'  is  equal 
to  the  circumference  of  the  circle. 

The  circumference  of  a  circle  may  be  obtained  in 
length  in  this  way.  Wrap  a  strip  of  paper  round  a 
circular  object,  e.  g.,  the  cylinder  in  Kindergarten 
gift  No.  JI.,  and  mark  off  two  coincident  points.  Un 
fold  the  paper  and  fold  through  the  points.  Then  the 
straignt  line  between  the  two  points  is  equal  to  the 
circumference  corresponding  to  the  diameter  of  the 
cylinder. 

By  proportion,  the  circumference  corresponding 
to  any  diameter  can  be  found  and  vice  versa. 


A'  D  G  A 

Fig.  87. 

Bisect  AA'  in  D  and  draw  DB  at  right  angles  to 
AA' ,  and  equal  to  the  diameter  of  the  generating 
circle. 

Then  A,  A'  and  B  are  points  on  the  curve. 

Find  O  the  middle  point  of  BD. 

Fold  a  number  of  radii  of  the  generating  circle 
through  O  dividing  the  semi-circumference  to  the 
right  into  equal  arcs,  say,  four. 

Divide  AD  into  the  same  number  of  equal  parts. 


IN  PAPER  FOLDING  147 

Through  the  ends  of  the  diameters  fold  lines  at 
right  angles  to  BD. 

Let  EFP  be  one  of  these  lines,  F  being  the  end  of 
a  radius,  and  let  G  be  the  corresponding  point  of  sec 
tion  of  AD,  commencing  from  D.  Mark  off  FP  equal 
to  GA  or  to  the  length  of  arc  BF. 

Then  P  is  a  point  on  the  curve. 

Other  points  corresponding  to  other  points  of  sec 
tion  of  AD  may  be  marked  in  the  same  way. 

The  curve  is  symmetric  to  the  axis  BD  and  corre 
sponding  points  on  the  other  half  of  the  curve  can  be 
marked  by  folding  on  BD. 

The  length  of  the  curve  is  4  times  BD  and  its  area 
3  times  the  area  of  the  generating  circle. 

THE  TROCHOID. 

281.  If  as  in   the   cycloid,   a   circle   rolls   along  a 
straight  line,  any  point  in  the  plane  of  the  circle  but 
not  on  its  circumference  traces  out  the  curve  called  a 
trochoid. 

THE  EPICYCLOID. 

282.  An  epicycloid  is  the  path  described  by  a  point 
on  the  circumference  of  a  circle  which  rolls  on  the 
circumference   of  another  fixed  circle  touching  it  on 
the  outside. 

THE  HYPOCYCLOID. 

283.  If  the  rolling  circle  touches  the  inside  of  the 
fixed  circle,  the  curve  traced  by  a  point  on   the  cir 
cumference  of  the  former  is  a  hypocycloid. 


148  GEOMETRIC  EXERCISES 

When  the  radius  of  the  rolling  circle  is  a  sub- 
multiple  of  the  fixed  circle,  the  circumference  of  the 
latter  has  to  be  divided  in  the  same  ratio. 

These  sections  being  divided  into  a  number  of 
equal  parts,  the  position  of  the  center  of  the  rolling 
circle  and  of  the  generating  point  corresponding  to 
each  point  of  section  of  the  fixed  circle  can  be  found 
by  dividing  the  circumference  of  the  rolling  circle  into 
the  same  number  of  equal  parts. 

THE  QUADRATRIX.* 

284.  Let  OACB  be  a  square.     If  the  radius  OA  of 
a  circle  rotate  uniformly  round  the  center  O  from  the 
position  OA  through  a  right  angle  to  OB  and  if  in  the 
same  time  a  straight  line  drawn  perpendicular  to  OB 
move  uniformly  parallel  to   itself   from   the   position 
OA  to  BC ';   the  locus  of  their  intersection  will  be  the 
quadratrix. 

This  curve  was  invented  by  Hippias  of  Elis  (420 
B.  C.)  for  the  multisection  of  an  angle. 

If  P  and  P'  are  points  on  the  curve,  the  angles 
A  OP  and  A  OP'  are  to  one  another  as  the  ordinates 
of  the  respective  points. 

THE  SPIRAL  OF  ARCHIMEDES. 

285.  If  the  line  OA  revolve  uniformly  round  O  as 
center,  while  point  P  moves  uniformly  from  O  along 
OA,  then  the  point  P  will  describe  the  spiral  of  Archi 
medes. 

*  Beman  and  Smith's  translation  of  Klein's  Famous  Problems  of  Elemen- 
tary  Geometry,  p.  57. 


THE 
RELIGION   OF  SCIENCE  LIBRARY 


A  Choice  Collection  of  Well  Made  Books.    Reproduc 
tion  of  Standard  Treatises  in  the  Departments  of 
Philosophy,    Psychology,    Biology, 
Religion,  etc. 


The  Religion  of  Science.     By  Paul  Carus.     30c. 

Three  Introductory  Lectures  on  the  Science  of  Thought.     By  F. 

Max  Miiller.     30c. 
Three  Lectures  on  the  Science  of  Language.    By  F.  Max  Miiller. 

30c. 

The  Diseases  of  Personality.     By  Th.  Ribot.     30c. 
The  Psychology  of  Attention.     By  Th.  Ribot.     30c. 
The  Psychic  Life  of  Micro-Organisms.     By  Alfred  Binet.     30c. 
The  Nature  of  the  State.     By  Paul  Carus.     20c. 
On  Double  Consciousness.     By  Alfred  Binet.     25c. 
Fundamental  Problems.     By  Paul  Carus.     60c. 
Diseases  of  the  Will.     By  Th.  Ribot.     30c. 
On    the    Origin    of  Language,    The   Logos    Theory.     By   Ludwig 

Noitre.     20c. 

The  Free  Trade  Struggle  in  England.     By  M.  M.  Trumbull.    30c. 
The  Gospel  of  Buddha.     By  Paul  Carus.     40c. 
Primer  of  Philosophy.     By  Paul  Carus.     30c. 
On  Memory,  and  The  Specific  Energies  of  the  Nervous  System. 

By  Ewald  Hering.     20c. 

An  Examination  of  Weismannism.     By  George  J.  Romanes.    40c. 
On  Germinal  Selection  as  a  Source  of  Definite  Variation.     By 

August  Weismann.     30c. 

Popular  Scientific  Lectures.     By  Ernst  Mach.     60c. 
Ancient  India;  Its  Language  and  Religions.     By  H.  Oldenberg. 

30c. 

The  Prophets  of  Israel.     By  C.  H.  Cornill.     30c. 
Thoughts  on  Religion.     By  G.  J.  Romanes.     60c. 
Philosophy  of  Ancient  India.     By  Richard  Garbe.     30c. 
Martin  Luther.     By  Gustav  Freytag.     30c. 
Rationalism.     By  George  Jacob  Holyoake.     30c. 
Chinese  Philosophy.     By  Paul  Carus.     30c. 
A    Mechanico-Physiological    Theory    of    Organic   Evolution.      By 

Carl  von  Nageli.      Summary.     30c. 
Mathematical   Essays   and  Recreations.     By   Herman   Schubert. 

30c. 

Truth  on  Trial.     By  Paul  Carus.     60c. 
The  Pleroma;  An  Essay  on  the  Sources  of  Christianity.    By  Paul 

Carus.     60c. 
The  Ethical  Problem.     By  Paul  Carus.     60c. 


THE 
RELIGION  OF  SCIENCE  LIBRARY 


Buddhism  and  Its  Christian  Critics.     By  Paul  Cams.     COc. 

Discourse  on  the  Method  of  Rightly  Conducting  the  Reason  and 
Seeking  Truth  in  the  Science.  By  Rene"  Descartes.  30c. 

Kant  and  Spencer.     By  Paul  Carus.     25c. 

The  Soul  of  Man.     By  Paul  Carus.     85c. 

Whence  and  Whither.     By  Paul  Carus.     35c. 

Enquiry  Concerning  the  Human  Understanding  and  Selections 
from  a  Treatise  of  Human  Nature.  By  David  Hume.  Paper, 
40c. 

An  Enquiry  Concerning  the  Principles  of  Morals.  By  David 
Hume.  30c. 

The  Psychology  of  Reasoning.         By  Alfred  Binet.     30c. 

A  Treatise  Concerning  the  Principles  of  Human  Knowledge.  By 
George  Berkeley.  30c. 

Three  Dialogues  Between  Hylas  and  Philonous.  By  George  Ber 
keley.  30c. 

Public  Worships  A  Study  of  the  Psychology  of  Religion.  By 
John  P.  Hylan.  30c. 

The  Meditations  and  Selections  from  the  Principles  of  Rene 
Descartes.  Tr.  by  John  Veitch.  40c. 

Leibniz's  Discourse  on  Metaphysics.  Tr.  by  Geo.  R.  Mont 
gomery.  60c. 

Kant's  Prolegomena.     Edited  in  English  by  Paul  Carus.     60c. 

St.  Anselms:  Proslogium,  Monologium,  an  Appendix  in  Behalf 
of  the  Fool  by  Oaunilon;  and  Cur  Deus  Homo.  Tr.  by  Sid 
ney  Norton  Deane.  60c. 

The  Metaphysical  System  of  Hobbes.  By  Mary  Whiton  Calkins. 
50c. 

Locke's  Essay  Concerning  Human  Understanding.  Books  II  and 
IV  (with  omissions).  By  Mary  Whiton  Calkins.  60c. 

The  Principles  of  Descartes'  Philosophy.  By  Benedictus  De 
Spinoza.  Paper,  40c. 

The  Vocation  of  Man.     By  Johann  Gottlieb  Fichte.     Paper,  30c. 

Aristotle  on  His  Predecessors.     Tr.  by  A.  E.  Taylor.     40c. 

Spinoza's  Short  Treatise  on  God,  Man  and  Human  Welfare.  Tr. 
by  Lydia  Gillingham  Robinson.  50c. 

The  Mechanistic  Principle.     By  Paul  Carus.     60c. 

Behind  the  Scenes  with  Mediums.     By  Abbot.     60c. 


Sent  postpaid  to  any  address  in  the  U.  P.  U.  at  prices  quoted. 
The  Open  Court  Pub.  Co.,  P.  O.  Drawer  F,  Chicago 


-i^-DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

ASTRON-MATH-STAT.  LIBRARY 

Tel.  No.  642-3381 

This  book  is  due  before  Library  closes  on  the  last  date 
stamped  below,  or  on  the  date  to  which  renewed. 
Ren  wed  books  are  subject  to  immediate  /recall. 


UCB 


Due  end  of  FALL 
Subject  to  rscall  a 


i/IATH  LIBRARY 


JSto- 


ter- 


NOV  1  4  1 


1  6  2005 


JAN    31989 


DtdWflOt  SUMMER  «omoGla 
Subject  to  recall  after- 


MAY  1  0  1996 


LD21-2Jt»-2'75 
(S4015slO)476—  A-32 


General  Library 

University  of  California 

Berkeley 


GENERAL  LIBRARY  -  U.C.  BERKELEY 


8000515531 


fftATt* 


